Simple Harmonic Motion

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Description of SHM’s

SHM spring pendulum, simple pendulum, physical pendulum SHM Other systems

Syllabus for IIT-JEE

Simple Harmonic Motion: Linear and Angular Simple Harmonic Motions. Introduction

Periodic Motion

A motion which repeats itself after equal intervals of time is called a periodic motion. In a periodic motion, force is always directed towards a fixed point which may or may not be on the path of motion. Examples of periodic motion:

• Circular motion is a periodic motion.

• Rotation of earth about her own axis is periodic. • Motion of a pendulum is periodic motion.

Any motion which repeats itself after regular interval of time is called periodic or harmonic motion

and the time interval after which the motion is repeated is called time period. A periodic motion can be either rectilinear or closed or open curvilinear (Fig. 1)

Motion of hands of a clock

(A)

Motion of earth around the sun

(B) Motion of a piston in a cylinder (C) Motion of a ball in a bowl (D) Motion of liquid in a U-tube (E) Some Periodic Motions:

Fig. 1 Oscillatory Motion:

A particle is said to be in oscillatory motion or vibratory if it moves periodically back and forth about a stable equilibrium position. Motion of pendulum is oscillatory. Oscillatory motion is a periodic motion between two fixed limits. Every oscillatory motion is a periodic motion but every periodic motion is not always oscillatory. Oscillatory or vibratory motion is a constrained periodic motion between certain precisely fixed limits. The world is full of oscillatory motions. The motion of the balance wheel of a watch, the motion of a piston in an automobile engine, the motion of the needle of sewing machine or the motion of a ball in a bowl are all oscillatory motions. The motion of molecules in a solid around fixed lattice sites, the perpetual motion of atoms within molecules, vibrations of the prongs of a tuning fork and the motion of electric and magnetic fields in an electromagnetic wave are few other examples of vibratory motion.

Note: Every oscillatory motion is periodic but every periodic motion is not oscillatory, e.g., the uniform circular motion of a conical pendulum or the motion of electrons around the nucleus in an atom are periodic but not oscillatory while the motion of a liquid in a U- tube or that of the free end of a clamped strip are oscillatory and hence periodic.

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Some oscillatory motions: Fig. 2

Simple Harmonic Motion:

If in case of oscillatory motion a particle moves back and forth (or up and down) about a fixed point (called equilibrium position or mean position) through a force or torque (called restoring force or torque) which is directly proportional to the displacement but opposite in direction, the motion is called simple harmonic and abbreviated as SHM. Further if the displacement in case of SHM is linear, the motion is called ‘Linear SHM’ and if the displacement is angular it is known as angular SHM. The up and down motion of a partially submerged floating body or of a mass attached to a spring are examples of linear SHM while the motion of a simple pendulum or physical pendulum or torsional pendulum (twisting and untwisting motion of a body suspended from a wire) are examples of angular SHM. T T cos 2 L T g T S T 2 m k S T 2 L g S T 2 I mgL S T 2 I C S T 2 1 c S Different types of pendulum

Fig. 3 12.1 SIMPLE HARMONIC MOTION:

When a body repeats its motion after regular time intervals we say that it is in harmonic motion or periodic motion. The time interval after which the motion is repeated is called the time period. If a body moves to and fro on the same path, it is said to perform oscillations. Simple harmonic motion

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(SHM) is a special type of oscillation in which the particle oscillates on a straight line, the acceleration of the particle is always directed towards a fixed point on the line and its magnitude is proportional to the displacement of the particle from this point. This fixed point is called the centre of oscillation. Taking this point as the origin and the line of motion as the X-axis, we can write the defining equation of a simple harmonic motion as

2 x

D Z … (1)

Where _Z2_{is a positive constant. If x is positive, a is negative and if x is negative, a is positive. This}

means that the acceleration is always directed towards the center of oscillation. If we are looking at the motion from an inertial frame

a = F/m.

The defining equation (1) may thus be written as F/m = - _Z2_x

Or, F = - m_Z2_x

Or, F = - kx. … (2)

We can use equation (2) as the definition of SHM. A particle moving on a straight line executes simple harmonic motion if the resultant force acting on it is directed towards a fixed point on the line and is proportional to the displacement of the particle from this fixed point. The constant 2

k mZ is called

the force constant or spring constant. The resultant force on the particle is zero when it is at the centre of oscillation. The centre of oscillation is therefore, the equilibrium position. A force which takes the particle back towards the equilibrium position is called a restoring force. Equation (2) represents a restoring force which is linear. Figure (4) shows the linear restoring force graphically.

Example 12.1

The resultant force acting on a particle executing simple harmonic motion is 4 N when it is 5 cm away from the centre of oscillation. Find the spring constant.

Solution:

The simple harmonic motion is defined as F = - k x.

The spring constant is k = F

x = 2 4 4 80 / . 5 5 10 N N N m cm m u .

12.2 QUALITATIVE NATURE OF SIMPLE HARMONIC MOTION:

Let us consider a small block of mass m in placed on a smooth horizontal surface and attached to a fixed wall through a spring as shown in figure (12.2). Let the spring constant of the spring be k.

m k

0 0 P

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The block is at a position O when the spring is at its natural length. Suppose the block is taken to a point P stretching the spring by the distance OP = A and is released from there.

At any point on its path the displacement x of the particle is equal to the extension of the spring from its natural length. The resultant force on the particle is given by F = - kx and hence by definition the motion of the block is simple harmonic.

When the block is released from P, the force acts towards the center O. The block is accelerated in that direction. The force continues to act towards O until the block reaches O. The speed thus increases all the time from P to O. When the block reaches O, its speed is maximum and it is going towards left. As it moves towards left from O, the spring becomes compressed. The spring pushes the block towards right and hence its speed decreases . The block moves to a point Q when its speed becomes zero. The potential energy of the system (block + spring), when the block is at P, is 1 2

2 k OP and when the block is at Q, it is 1 2

2 k OQ . Since the block is at rest at P as well as at Q, the kinetic energy is zero at both these positions. As we have assumed frictionless surface, principle of conservation of energy

gives 1 2 1 2

2 k OP 2 k OQ or OP = OQ.

The spring is now compressed and hence it pushes the block towards right. The block starts moving towards right, its speed increases up to O and then decreases to zero when it reaches P. Thus, the particle oscillates between P and Q. As OP = OQ, it moves through equal distances on both sides of the centre of oscillation. The maximum displacement on either side from the center of oscillation is called the amplitude.

Example 12.2

A particle of mass 0.50 kg executes a simple harmonic motion under a force F = –(50 N/m)x. If it crosses the center of oscillation with a speed of 10 m/s, find the amplitude of the motion.

Solution:

The kinetic energy of the particle when it is at the center of oscillation is

2

1 2

E mv 1 0.05 10 / 2

2 kg m s = 25 J.

The potential energy is zero here. At the maximum displacement x = A, the speed is zero and hence the kinetic energy is zero. The potential energy here is 1 2

2k A . As there is no loss of energy,,

2

1

25

2kA J … (i)

The force on the particle is given by F = - (50 N/m)x. Thus, the spring constant is k = 50 N/m.

Equation (i) gives 1 50 / 2 25 .

2 N m A J or A = 1 m.

12.3 EQUATON OF MOTION OF A SIMPLE HARMONIC MOTION:

Consider a particle of mass m moving along the X – axis. Suppose, a force F = – kx acts on the particle where k is a positive constant and x is the displacement of the particle from the assumed origin. The particle then executes a simple harmonic motion with the center of oscillation at the origin. We shall calculate the displacement x and the velocity v as a function of time.

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Suppose the position of the particle at t = 0 is x₀ and its velocity is v₀. Thus, At t = 0, x = x₀ and v = v₀.

The acceleration of the particle at any instant is F k x 2 x

m m D Z Where Z k m/ . Thus, dv 2x dt Z … (12.3) Or dv dx 2x dx dt Z Or, v dv 2x dx Z Or, vdv Z2x dx.

The velocity of the particle is v₀ when the particle is at x = x₀. It becomes v when the displacement becomes x. We can integrate the above equation and write

0 0 2 . v x v x v dv Z x dx

³

or, 0 0 2 2 2 2 2 v x v x v _Z x ª º ª º « » « » « » « » ¬ ¼ ¬ ¼ or, v2v02 Z2 x2x02 . or, 2 2 2 2 2 2 0 0 v v Z x Z x or, 2 2 2 2 2 0 0 v v Z x Z x or, 2 2 0 2 v v Z x Z § · ¨ ¸ ¨ ¸ © ¹ 2 2 2 0 0 2 v x x Z § · ¨ ¸ © ¹ Writing 2 2 2 0 0 v x A Z § · ¨ ¸ © ¹ … (12.4)

The above equation becomes 2 2

v Z A x … (12.5)

We can write this equation as dx A2 x2

dt Z or, 2 2 . dx dt A x Z

At time t = 0 the displacement is x = x₀ and at time t the displacement becomes x. The above equation

can be integrated as x t dx dt Z

³

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or, 0 1 0 sin x t x x t A Z ª _º « » ¬ ¼ or, 1 1 0 sin x sin x t A A Z Writing sin 1x0 A G _{, this becomes} 1 sin x t A Z G or, x = A sin ZtG … (12.6) The velocity at time t is v dx A cos t

dt Z Z G . … (12.7)

12.4 TERMS ASSOCIATED WITH SIMPLE HARMONIC MOTION: (a) Amplitude:

Equation (12.6) gives the displacement of a particle in simple harmonic motion. As sin ZtG can take values between – 1 and + 1, the displacement x can take values between – A and + A. This gives the physical significance of the constant A. It is the maximum displacement of the particle from the center of oscillation, i.e., the amplitude of oscillation.

(b) Time Period:

A particle in simple harmonic motion repeats its motion after a regular time interval. Suppose the particle is at a position x and its velocity is v at a certain time t. After some time the position of the particle will again be x and its velocity will again be v in the same direction. This part of the motion is called one complete oscillation and the time taken in one complete oscillation is called the time period. T. Thus, in figure (12.4) Q to P and then back to Q is a complete oscillation, R to P to Q to R is a complete oscillation, O to P to Q to O is a complete oscillation etc. Both the position and the velocity (magnitude as well as direction) repeat after each complete oscillation.

We have, x = A sin ZtG .

If T be the time period, x should have same value at t and t + T. Thus, sin ZtG sin[Z tT G].

Now the velocity is (equation 12.7) v = AZcos ZtG

As the velocity also repeats its value after a time period, cos ZtG cos[ (Z tT)G].

Both sin ZtG and cos ZtG will repeat their values if the angle ZtG increases by 2Sor its multiple. As T is the smallest time for repetition.

2 t T t Z G Z G S or, ZT 2S or, T 2S Z .

Remembering that Z k m/ , we can write for the time period. 2 2 m T k S S Z ... (12.8)

Where k is the force constant and m is the mass of the particle.

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Example 12.3

A particle of mass 200 g executes a simple harmonic motion. The restoring force is provided by a spring of spring constant 80 N/m. Find the time period.

Solution

The time period isT 2 m

k S 3 200 10 2 80 / kg N m S u 2Su0.05s 0.31 .s

(c) Frequency and Angular Frequency:

The reciprocal of time period is called the frequency. Physically, the frequency represents the number of oscillations per unit time. It is measured in cycles per second also known as hertz and written in symbols as Hz. Equation (12.8) shows that the frequency is

1 2 v T Z S ... (12.9) 1 2 k m S ... (12.10).

The constant Zis called the angular frequency.. (d) Phase:

The quantity I Zt is called the phase. It determines the status of the particle in simple harmonicG

motion. If the phase zero at a certain instant, x = A sin ZtG = 0 and v = A Zcos ZtG = A I Z. This means that the particle is crossing the mean position and is going towards the positive direction. If the phase is S/ 2, we get x = A, v = 0 so that the particle is at the positive extreme position. Figure (12.5) shows the status of the particle at different phases.

We see that as time increases the phase increases. An increases of 2S brings the particle to the same status in the motion. Thus, a phase Zt is equivalent to a phaseG Zt G 2S. Similarly a phase change of 4 , 6 , 8S S S ………… etc, are equivalent to no phase change.

Figure (12.6) shows graphically the variation of position and velocity as a function of the phase.

S S I S S I

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(e) Phase constant:

The constant G appearing in equation (12.6) is called the phase constant. This constant depends on the choice of the instant t = 0. To describe the motion quantitatively, a particular instant should be called t = 0 and measurement of time should be made from this instant. This instant may be chosen according to the convenience of the problem. Suppose we choose t = 0 at an instant when the particle is passing through its mean position and is going towards the positive direction. The phase tZ should thenG be zero. As t = 0 this means G will be zero. The equation for displacement can then be written as

x = – A sin tZ .

If we choose t = 0 at an instant when the particle is at its positive extreme position, the phase is S/ 2at this instant. Thus Z G St / 2 and hence G S/ 2. The equation for the displacement is x = AA sinZ St / 2. or, x = A cos tZ .

Any instant can be chosen as t = 0 and hence the phase constant can be chosen arbitrarily. Quite often we shall choose G and write the equation for displacement as x = A sin t0 Z . Sometimes we may have to consider two or more simple harmonic motions together. The phase constant of any one can be chosen as G . The phase constants of the rest will be determined by the actual situation. The0 general equation for displacement may be written as

x = A sin ZtG sin '

2

A §_¨ZtS G ·_¸

Where 'G is another constant. The sine form and the cosine form are, therefore, equivalent. The value of phase constant, however, depends on the form chosen.

Example 12.4:

A particle execute simple harmonic motion of amplitude Along the X –axis. At t = 0, the position of the particle is x = A/2 and it moves along the positive x – direction. Find the phase constant G if the equation is written as x Asin ZtG .

Solution:

We have x Asin ZtG . At t = 0, x = A/2. Thus, A/2 = A sin G or, sinG =1/2 or, , G = S / 6 or 5. The velocity is v dx A cos t .

dt Z Z G At t = 0, v = A ZcosG . Now 3 5 3 cos cos 6 2 and 6 2 S S

As v is positive at t = 0, G must be equal to S/ 6

12.5 SIMPLE HARMONIC MOTION AS A PROJECTION OF CIRCULAR MOTION:

Consider a particle P moving on a circle of radius A with a constant angular speed Z (figure 12.7). Let us take the center of the circle as the origin and two perpendicular diameters as the X and Y – axes. Suppose the particle P is on the X – axis at t = 0. The radius OP will make an angle T Zt with the X – axis at time t. Drop perpendicular PQ on X – axis and PR on Y – axis. The x and y – coordinates of the particle at time t are

x = OQ = OP cos tZ

or, x = A cos tZ … (12.11)

And y = OR = OP sin tZ

or, y = A sin tZ . ... (12.12)

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Equation (12.11) shows that the foot of perpendicular Q executes a simple harmonic motion on the X – axis. The amplitude is A and the angular frequency is Z. Similarly, equation (12.12) shows that the foot of perpendicular R executes a simple harmonic motion on the Y – axis. The amplitude is A and the angular frequency is Z. The phases of the two simple harmonic motions differ by S/ 2 (remember

cosZt sin(ZtS / 2)).

Thus, the projection of a uniform circular motion on a diameter of the circle is a simple harmonic motion.

12.6 ENERGY CONSERVATION IN SIMPLE HARMONIC MOTION: Simple harmonic motion is defined by the equation F = - kx.

The work done by the force F during a displacement from x to x + dx is dW = F dx = – kx dx the work done in a displacement from x = 0 to x is

2 0 1 . 2 x W

³

kx dx kx

Let U (x) be the potential energy of the system when the displacement is x. As the change in potential energy corresponding to a force is negative of the work done by this force

2

1 0

2

U x U W kx .

Let us choose the potential energy to be zero when the particle is at the center of oscillation x = 0.

Then 0 0 1 2

2

U and U x k x .

This expression for potential energy is same as that for a spring and has been used so far in this chapter. As k ,k m 2 m Z Z We can write 1 2 2 2 U x mZ x ... (12.13)

The displacement and the velocity of a particle executing a simple harmonic motion are given by sin

x A ZtG and v AZcos ZtG .

The potential energy at time t is, therefore, 1 2 2 2 U mZ x 2 2 2 1 sin . mZ A ZtG

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And the kinetic energy at time t is 1 2 2

K m v 1 2 2 cos2 .

2mZ A ZtG The total mechanical energy at time t is E = U + K

2 2 2 2 1 sin cos 2mZ A Zt G Zt G ª º ¬ ¼ 2 2 1 . 2mZ A … (12.14)

We see that the total mechanical energy at time t is independent of t. Thus, the mechanical energy remains constant as expected.

As an example, consider a small block of mass m placed n a smooth horizontal surface and attached to a fixed wall through a spring constant k (figure 12.8).

When displaced from the mean position (where the spring has its natural length), the block executes a simple harmonic motion. The spring is the agency exerting a force F = - kx on the block. The potential energy of the system is the elastic potential energy stored in the spring.

At the mean position x = 0, the potential energy is zero. The kinetic energy is 02 2 2

1 1

. 2m v 2mZ A All the mechanical energy is in the form of kinetic energy here. As the particle is displaced away from the mean position, the kinetic energy decreases and the potential energy increases. At the extreme positions x = r A, the speed v is zero and the kinetic energy decreases to zero. The potential energy is increased to its maximum value 1 2 1 2 2.

2k A 2mZ A All the mechanical energy is in the form of potential energy here.

Example: 12.5

A particle of mass 40 g executes a simple harmonic motion of amplitude 2.0 cm. If the time period is 0.20 s, find the total mechanical energy of the system.

Solution:

The total mechanical energy of the system is 1 2 2 2 E mZ A = 2 2 2 2 1 2 2 2 m A m A T T S S § · ¨ ¸ © ¹ = 2 2 2 2 2 2 40 10 2.0 10 0.20 kg m s S _u _u 2 7.9 10u J.

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12.7 ANGULAR SIMPLE HARMONIC MOTION:

A body free to rotate about a given axis can make angular oscillations. For example, a hanging umbrella makes angular oscillations when it is slightly pushed aside and released. The angular oscillations are called angular simple harmonic motion if.

(a) There is a position of the body where the resultant torque on the body is zero, this position is the mean positionT .0

(b) When the body is displaced through an angle from the mean position, a resultant torque acts which is proportional to the angle displaced, and

(c) This torque has a sense (clockwise or anticlockwise) so as to bring the body towards the mean position.

If the angular displacement of the body at an instant isT, the resultant torque acting on the body in angular simple harmonic motion should be T = - kT.

If the moment of inertia is I, the angular acceleration is T k

I I D .T or, 2 2 2 d dt T Z T … (12.15) Where, Z k I/

Equation (12.15) is indentical to equation (12.3) except for the symbols. The linear displacement x in (12.3) is replaced here by the angular displacementT. Thus, equation (12.15) may be integrated in the similar manner and we shall get an equation similar to (12.6), i.e.,

0 sin t

T T Z G … (12.6)

Where T0 is the maximum angular displacement on either side. The angular velocity at time t is given

by, 0 cos . d t d t T _{T Z} _Z _G : _{… (12.17)}

The time period of oscillation is 2 2 I T k S S Z … (12.18)

And the frequency of oscillation is

1 1

2

k v

T S I … (12.19)

The quantity Z k I/ is the angular frequency.. Example: 12.6

A body makes angular simple harmonic motion of amplitude S/ 10 rad and time period 0.05 s. If the body is at a displacement T S/ 10rad at t = 0, write the equation giving the angular displacement as a function of time.

Solution:

Let the required equation be T T0sin ZtG .

Here T₀ = amplitude 10

S rad

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1 2 2 40 0.05 s T s S S Z S So that sin 40 1 10rad s t S T § · _ª S G_º ¨ ¸ _¬ _¼ © ¹ . … (i)

At t = 0, T S/ 10rad. Putting in (i) sin 10 10 S S G § · ¨ ¸_© _¹

or, sinG 1 or, G S/ 2. Thus by (i), 1 sin (40 ) 10rad s t 2 S S T § · ª S º ¨ ¸ « » © ¹ ¬ ¼ = 1 cos (40 ) 10rad s t S S § · _ª _º ¨ ¸ _¬ _¼ © ¹ .

Energy:-The potential energy is 1 2 1 2 2

2 2

U kT IZ T and the kinetic energy is 1 2 2

K I: .

The total energy is E = U + K = 1 2 2 1 2. 2IZ T 2I: Using T T0 sin ZtG 2 2 2 0 1 sin 2 E IZ T ZtG 02 2 2 1 cos 2IT Z Zt G 2 2 0 1 . 2IZ T … (12.20).

Composition of two ‘parallel’ SHM’s

We have already seen that, for a particle performing SHM, the restoring force is given by, F = - kx. Now, if the particle is acted upon by two restoring forces along same line instead of one, then the

resultant motion can be studied using superposition principle. Hence, if F

₁ and F₂ are the two forces acting on the particle, then using superposition principle, we first assume that only the force F₁ is acting on the particle and find the corresponding position of the particle as,

x₁ = A₁ sin ZtI1 ... (73)

Then we assume that only the force F₂ acts on the particle and find the corresponding position of the particle as,

x₂ = A₂ sin ZtI2 ... (74)

Here, we assume the frequency of the oscillation to be same. The resultant position of the particle is then given by,

x = x₁ + x₂ ... (75)

1sin 1 2 sin 2

x A Zt I A Zt I

?

1{sin .cos 1 cos .sin }1 2{sin .cos 2 cos .sin 2}

x A Zt I Zt I A Zt I Zt I

?

1 1 2 2 1 1 2 2

sinZt A[ cosI A cosI ] cos Zt A[ sinI A sinI ]

Let, A1cosI1A2cosI2 = X … (76)

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And A1sinI1A2sinI2 = YY … (77)

? x XsinZtYcosZt

Let, X2_{+ Y}2_{= A}2 _{… (78)}

We can write,

cos & sin tan X A Y A Y X T T T ½ ° ¾ _°¿ … (79)

sin cos cos sin

x A Zt T A Zt T

?

sin

x A Zt T

? ... (80)

Equation (80) shows that the resultant of two SHM’s along the same path is itself a SHM. The amplitude of this resultant SHM is A and it’s relation with A₁ and A₂ is given by,

2 2

1 2 2 1 2cos 1 2

A A A A A I I ... (81)

This relation can be easily derived using equations (76) to (78). Also, the initial phase T, is given by,,

1 1 2 2 1 1 2 2 sin sin tan cos cos A A Y X A A I I T I I ... (82)

The difference I I1 2in equation (81) is called the phase difference between the two motions.

If I1 I2, then the two simple harmonic motions are in phase and we get, A = AA1 + A2 &

1 2

tanT tanI tanI . In this case, the amplitude is the maximum possible amplitude.

If I I₁ ₂ S then the two SHM’s are out of phase and we get, A A1A and2 tanT tanI2

If A₁ = A₂, then A = 0 and the particle does not oscillate.

For any value of I I1 2 other than 0 and S the resultant amplitude is between A1A2 and AA₁ + AA₂.

Vector Combination of Two Simple Harmonic Motion: Suppose the two individual SHM s are represented by

1 1 sin

x A Zt

2 2 sin

x A ZtG .

The two SHM s can be represented by vectors with magnitude A₁ and A₂_{respectively. The resultant A})& of these two vectors represent the resultant SHM.

2 2 1 2 1 2 |A| A A 2A A cosG )& and 2 1 2 sin tan cos A A A G G

Case: If the two SHM’s are perpendicular, then

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1 sin x A Zt ... (1) 2 sin y A ZtG ... (2) By equation (1), 2 2 1 1 sin, x , cos t 1 x A Z A

By equation (2), y A2 sinZtcosGcosZtsinG

2 2 2 1 1 cos sin 1 x x A A G G A ª º « » « » ¬ ¼ 2 ₂ 2 2 2 1 1 cos 1 sin y y x A A G A G § · § · ¨ ¸ ¨_¨ ¸_¸ © ¹ © ¹ 2 2 2 2 2 2 2 2 2 1 2 2 1 1 2

cos cos sin sin

y x y x x A A A G A G G A G 2 2 2 2 2 1 2 1 2 2 cos sin x y x y A A A A G G

The is equation of an ellipse and hence the particle moves in ellipse. Here note that the above cases are valid only when the frequencies of the two SHMs are equal.

15.3 A Few Graphs Concerning SHM (a) Graph of Displacement Versus Time

Assuming the displacement equation to be x = a sin Zt, clearly the graph will be a sine curve. (Fig 15.3)

Putting x = 0, we have sin (Zt) = 0

, n n t n where n I t S Z S Z

Putting n = 0, 1, 2, 3 etc, we get

0 1 2 3

2 3

0, , ,

t t S t S t S

Z Z Z etc, and putting x = r a (i.e., maximum displacement) we get sin 1 t Z r . 2 1 2 n t n S where n I Z .

Putting n = 0, 1, 2 etc, we get.

0 1 2 3 5 , , 2 2 2 t S t S t S Z Z Z , etc

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(b) Graph of Velocity versus Time.

Assuming x = a sin Zt, we get on differentiating cos

dx

v a t

dt Z Z .

Putting v = 0; we get cos Zt= 0 2 1 ; 2 n t n S where n I Z Putting n = 0, 1, 2 etc. 0 1 3 3 5 , , 2 2 2 t S t S t S Z Z Z etc.

Putting v = ar Z (velocity amplitude), we get cos Zt r .1 n t n where n I Z S , n n t S Z

_{putting n = 0, 1, 2 etc, we get} 0 1 2

2 0 : ; . t t S t S etc Z Z . 2 / O x time (t) 3 /2 / /2

The graph is a cosine curve (Fig. 15.4) (c) Graph of Acceleration Versus Time:

Assuming x = a sin Zt, we have f = - Z2x Z2asinZt

Putting f = 0, the instants at which f = 0, can be found; sin Zt= 0. n t n Z S Putting n = 0,1, 2,……….. 0 1 2 2 0, , t t S t S Z Z ? _etc.

Putting f = _#_Z2_{a we have sin}Z_t_{r (acceleration amplitude).}₁

2 1 2 n t n S Z r Putting n = 0, 1, 2, etc. 0 1 2 3 5 , , 2 2 2 t S t S t S Z Z Z, etc.

Clearly the graph is a sine curve (figure 15.5)

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(d) Graph of Acceleration Versus Displacement

Since acceleration (f) = - _Z2_{(x). The variation is linear (figure. 15.6)}

(e) Graph of Velocity Versus Displacement we have seen earlier that 2 2

v Z a x 2 2 2 2 v Z a x 2 2 2 2 2 2 2 2 2 1 v x v a x a a Z Z _.

The graph is an ellipse. (Figure. 15.7).

(f) Graph of Variation between Displacement (x) and Phase ZtI . From the equation x = a sin ZtI

x = 0 at ZtI = nS where n I . And 2 1 , 2

x ra at ZtI n S where n I .

? The graph is a sine curve (Fig. 15.8)

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(g) Graph of Velocity Versus Phase. From the equation v = aZcos ZtI .

v = 0 at 2 1

2

t n S where n I

Z I  . v raZ at ZtI nS where n I

The graph is shown in figure 15.9.

Example:1

A particle executes SHM between two points separated by 10 cm. If the force experienced by it be 2N at a displacement of 1 cm from the mean position, then find the maximum force experienced by it. Solution:

Evidently, the maximum displacement 10 5 2 cm.

For a particle executing SHM the force experienced is given by

F = - kx ... (1).

And F_max = - k (x_max) ... (2). Dividing (2) by (1) Fmax xmax

F x max max 5 10 . 2 1 F F N Example 2

A particle executing SHM oscillates between two fixed points separated by 20 cm. If its maximum velocity be 30 cm/s, find its velocity when its displacement is 5 cm from its mean position.

Solution:

Evidently, the max, displacement =20 10

2 cm

For a particle executing SHM the speed at a displacement ‘x’ is given by.

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2 2 v Z a x ... (1) and vmax Za ... (2) 2 2 max v a v _a _x 2 2 max 10 10 2 3 3 5 3 10 5 v v 2 3 30 20 3 3 v u cm/s. Example 3:

A particle executes SHM with an angular frequency Z 4S rad/sec. If it is at its extreme position initially, then find the instants, when it is at a distance 3

2 times, its amplitude from the mean position. Solution:

Let a be the amplitude and I the phase constant for the oscillation. The displacement equation can be written as sin 4 4 / sec x a StI Z Srad Since at t = 0; x = + a. 4 0 2 2 S S S I I ? .

Now, let t_n be the instant, when the particle’s displacement is 3 2 a , nth time (where n I ). Evidently 3 cos 4 2 a a a St r 2 1 4 6 4 24 n n n t n S t S S r r .

Taking only the positive values for ‘t’.

1 2 3 4 5 1 5 7 11 13 , , , , , . 24 24 24 24 24 t s t s t s t s t s etc Example 4

A particle executes SHM with amplitude 8 cm and a frequency 10 sec-1_{. Assuming the particle to be at}

a displacement 4 cm, initially, in the positive direction, determine its displacement equation and the maximum velocity and acceleration.

Solution

Given a = 8 cm and Z = 2Sn

= 20 S rad/sec. Let the phase constant be I.

The displacement equation can be written as x = 8 sin (20St +I) Given, at t = 0; x = 4 cm 4 8sin 20S 0 I ? 1 sin 2 6 S I I

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? The displacement equation becomes 8 sin 20 6

x §_¨ StS·_¸

Differentiating the above equation w.r.t time ‘t’ 160 cos 20 6 dx v t dt S S § S · _¨ _¸ © ¹ max 160 / [ cos 20 / 6) 1 ] v Scm s when St S ? r r .

Differentiating once again w. r. t time t.

2 2 2 3200 sin 20 / 6 d x f t d t S S S 2 max 3200 / f S cm s ? # [when sin (20StS/ 6) r ].1 Example 5

If the maximum speed and acceleration of a particle executing SHM be 20 cm/s and 100 S cm/s2_{, find}

the time period of oscillation. Solution

Given |vmax| aZ 20cm s/ … (1).

And | f_max| aZ2 100Scm s/ 2 … (2). Dividing equation (2) by (1) we get Z 5S rad/sec.

? Time period T 2_ZS 2₅_SS 0.4 sec. Example 6

If a particle executing SHM possesses a speed of 10 cm/s, when at mean position, find its speed when at a displacement of half the amplitude.

Solution

Let ‘a’ be the amplitude of oscillation. From velocity 2 2 v Z a x Squaring, we get v2 Z2 a2x2 Given at x = 0; v = 10 2 2 100 Z a ? … (1). And let at x = ; ' 2 a v v 2 2 2 3 4 a vc Z § · ¨_¨ ¸_¸ © ¹ … (2).

Dividing equation (2) by (1), we get

2 ' 3 10 4 v § · ¨ ¸ © ¹ 3 u

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Example 7

Two particles execute SHM with same frequency and amplitude along the same straight line. They cross each other, at a point midway between the man and the extreme position. Find the phase difference between them.

Solution

Let ‘a’ be the amplitude and I1andI2 their respective phases at any instant, when they cross each

other.

Then their displacement equations can be written as x = a sin (phase) for the two particles we have

1 2

sin sin

2 2

a a

a I and a I

Taking the least value, 1

6 S

I _{. And taking the next possible value} I₁ I₂ _.

2 5 6 S I Moreover, from v aZ & cos (phase) If v and v1 2 & &

be the respective velocities of the two particles, then v1 v2

& &

i.e., aZcos I1 aZcos I2 cosI2 cos S Ir 1

1 2 5 7 , 6 6 6 If I S thenI S or S ?

? Required phase difference 2 2

2 3

S

I I _rad.

Note: The phase difference can be in general 2 2 ;

3 n n I

S _S

Example: 8

A particle executes SHM with time period of 2 sec; find the time taken by it to move from one extreme position to the nearest mid point between mean position and extreme position.

Solution

Let ‘a’ be the amplitude of oscillation then evidently 2

T

S

Z S rad/sec. The displacement equation can be written as x asin StI

Let, for simplicity, the time be reckoned from the instant when the particle was as its extreme position. Thus, at t = 0; x = a

sin

2

a a I I S

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, sin 2 Thus x a §_¨StS·_¸ © ¹ Let at 1 2 a t t ve x 1 sin 2 2 a a §St S · ? _¨ _¸ © ¹ 1 1 5 2 6 3 t S S t S S S § · _¨ _¸ © ¹ 1 1 3 t _s. Example 9

If two SHMs are represented by

1 10 sin 4 2 5 [sin 2 8 cos 2 ]

2

y §_¨ StS·_¸and y St St

Solution

For the equation 1 10 sin 4

2

y §_¨ StS ·_¸

equationy2 5 [sin 2St 8cos 2St]. Multiplying and dividing by 1 8 3

2 1 8 15 sin 2 cos 2 3 3 y ª« St u St º» « »

¬ ¼ 15sin 2 tS I where tan I 8

Whose amplitude a₂ is obviously 15 units. ? The required ratio 1

2 10 2 15 3 a a . Example 10

A particle is executing SHM with amplitude ‘a’ and maximum velocity v₀. Find the displacement when its velocity is 3

2

v

, Solution

From the equation for speed

2 2 2 2 v Z a x 2 2 2 0 v Z a ... (1) And 2 2 2 2 0 3 4 v a x Z ... (2) Dividing (1) by (2), we get 2 2 2 2 2 4 3 1 3 4 a x or a x a 2 2 1 x x r .a

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Example 11

A small particle moves along a horizontal circle of radius 5 cm with a uniform angular speed of 6 S rad/sec. The circle stands in front of a vertical wall, whereon a normal beam of light casts a sharp shadow. What is the maximum speed and maximum acceleration of the shadow? Also find the range between whose limits the shadow moves.

Solution: The situation is similar to the one discussed above. The shadow executes SHM upon the wall. The maximum speed vmax Za 6Su 5 30Scm s/ and the maximum acceleration

2 max

f Z a

2 2 2

6S u 5 180S cm s/

The shadow moves along a straight line over the projection of circle upon the wall and its range is the diameter i.e., 10 cm.

Example 12

Find the position (s) of a particle executing SHM, when its K.E. is 1

4th its T.E. Solution

From equation (8) K = 1 2 2 2

2mZ a x and from equation (11) E =

2 2

1

2m a Z from the problem

4 E K or 4K = E 2 2 2 2 2 1 1 4 ( ) 2mZ a x 2mZ a ª º « » ¬ ¼ 4a24x2 a2 4x2 3a2 3 2 a x r . Example 13

Find the ratio of K.E. to P.E. for a particle executing SHM, when it is midway between the mean position and extreme position.

Solution Given 2 a x r Now 2 2 2 2 2 2 2 2 1 2 1 2 m a x K a x U _m _x x Z Z 2 2 2 4 ₃ 1 4 a a K U a § · ¨ ¸ ¨ ¸ © ¹ ? _. Example 14

What fraction of total energy is carried by a particle executing SHM, in the form of kinetic energy when the displacement of the particle is 1

4th the maximum displacement, from one of its extreme position?

Solution

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Displacement of the particle from its mean position is 3

4 4

a a

a (say)x

Now, fraction of T.E. in the form of K.E.

2 2 2 2 2 2 1 2 ₁ 1 2 m a x K x E a m a Z Z _{§ ·} _{u ¨ ¸} © ¹ 2 3 7 1 4 16 K E § · _{¨ ¸} © ¹ . Example 15

A particle of mass 200 g is executing SHM with an amplitude 20 cm. If its kinetic energy be 32×10–3

joule initially, determine its displacement equation, assuming the phase constant to be 4 S . Solution

Let Z be the angular frequency, for the SHM, then displacement equation can be written as x = 20 sin ZtS / 4 where t is in sec. and x is in cm.

Now, initially x 20sin§ ·¨ ¸_{© ¹}S₄ [Putting t = 0] 20 2 2 400 200 2 x

Now, kinetic energy at a displacement x, will be 1 2 2 2 2 K mZ a x 3 1 200 2 2 32 10 400 200 10 2 1000 Z § · u u_¨ _¸u u © ¹ 2 2 32 2 200 200 10 Z u u u 8 4 / sec 2 rad Z

Thus, the displacement equation will be 20 sin 4 4

x §_¨ tS ·_¸

Example 16

A particle executes SHM with force constant 2 × 106_{N/m and amplitude 1 cm, and has a total mechanical}

energy of 200 J. Find the following (a) maximum K.E., (b) Maximum P.E., (c) minimum K.E., (d) minimum P.E.

Solution

Given k mZ2 u2 10 ;6 a u1 102 m

(a) Maximum 2 2 max

1 . . 2 K E mZ a ª_¬v Zaº_¼ 6 4 1 2 10 1 10 100 2 J u u u u .

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(b) From E = K + U

max

E U

_{[when K = 0; at the extreme position]} ? Maximum potential energy U_max = 200 J.

(c) Minimum kinetic energy min 2

1

2m v = 0 [at the extreme portion] (d) Minimum Potential energy U_min = E - K_max = 200 – 100 = 100 J. Note: The particle’s potential energy is not zero at mean position

Example 31

A uniform rod of length l and mass m, is suspended by one end and is displaced slightly, to undergo angular SHM. Find its time period.

Solution

Consider the rod, in a position making an angle ' 'T with the vertical at any instant. (Fig. 15.43) Net torque acting on the rod about the point of suspension.

sin 2

l mg

W T m g l [for small anglesin ]

l

T

T T

_|

Now, angular acceleration

2 2 2 3 d I d t ml T W W 2 2 2 3 2 ml d m g l d t T T W ? _or, 2 2 3 0 2 d g l d t T _T . Comparing with the standard form

2 2 2 0 d d t T Z T We get, 3 2 g l Z _. ? The period T 2_ZS 2S ₃2_gl . Example 32

A square lamina of edge ‘a’ is suspended through a hole punched on a diagonal at a distance of _{2 2}

a

from its center. Find the period of oscillations.

Solution: Consider, the oscillating lamina (of mass m say) at any position making an angle ' 'T with the vertical. Net torque acting on the lamina about the point of suspension O, will be (Figure. 15.44)

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sin

2 2 2 2

a a

m g mg

W T| T

Now, the moment of inertia of the lamina about an axis through O, and normal to its plane will be.

2 2 2 6 2 2 ma a I ¨m§_¨ ·¸_¸ © ¹ 2 7 24 ma Since, 2 2 I d dt T W 2 2 2 7 24 2 2 ma d m g a dt T T § · ¨ ¸ ¨ ¸ © ¹ 2 2 (12) 7 2 d g dt T T

Comparing with the standard form

2 2 2 d dt T Z T We get 2 12 7 2 g Z _. ? The period 2 2 7 2 7 2 12 3 T g g S S S Z

15.11 Average Values of a Few Parameters:

Average velocity and acceleration of a particle executing SHM.

The average of any time dependent function (Z = f(t)) say) can be obtained, between the instants t = t₁ to t = t₂. As 2 2 1 1 2 2 1 1 t t t t t t t t Zdt f t dt Z dt dt !

³

.

(a) Average Velocity: Since the motion in a SHM, repeats again and again over a period of one time period, therefore the average velocity over one complete time period is zero. However, the average velocity between one extreme position to the other will not be zero. This can be calculated as follows.

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O V

t

Assuming the equation for velocity in terms of time as v aZcos Zt 3 / 4 / 4 3 / 4 / 4 cos T T T T a t d t v d t Z Z ? !

³

3 / 4 / 3 / 4 / 4 3 sin sin sin 4 4 3 4 4 T T A T T T T t a a T T t Z Z Z Z Z ª § · § ·º ª º ¨ ¸ ¨ ¸ « » « » _© _¹ _© _¹ ¬ ¼ ¬ ¼ § · ¨ ¸ © ¹ 2 3 sin sin [ 2 ] 2 2 a T T S S Z S § · § ·½ ® ¨_© ¸_¹ ¨ ¸_{© ¹}¾ ¯ ¿ 2 {( 1) (1)} a T 2 4a a v T Z S ? ! 2

S times maximum velocity ... (24).

Alternatively: Average velocity over one half oscillation = Average speed from one extreme position to another. [ there is no variation in direction]

tan cov 2 4

/ 2

total dis ce ered a a

total time taken T T .

(b) Average Acceleration: Arguing in the same way as for velocity, the average acceleration over one complete oscillation will be zero. However if we are not interested in one complete oscillation, it may not turn out to be zero.

Let us find out the average acceleration between the instants when the particle crosses its mean position two consequtive times.

Assuming the equation for acceleration as f Z2asinZt

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/ 2 2 0 / 2 0 sin T T a t d t f d t Z Z !

³

/ 2 2 0 2 cos t T a T Z Z Z § · ª º _{¨ ¸ «} _» © ¹ ¬ ¼ 2 2 cos cos 0 2 a T T Z Z Z ª § · º « ¨ ¸ » © ¹ ¬ ¼ 2 2 ₂ 2 cos 0 a a T Z Z S Z ª¬ º¼ S . 2 2 ₂ , | | a or f Z S S

! time maximum acceleration. … (25)

Average Kinetic Energy, Potential Energy and Total energy. (a) Average kinetic energy over one complete oscillation.

The instantaneous K.E. of a particle executing SHM is given by 1 2 2cos2 2 K mZ a Zt 2 2 2 0 0 1 cos ( ) 2 T T m a t dt K d t Z Z ? !

³

2 2 0 sin 2 1 4 2 r t m a t T Z Z Z ª º « » « » ¬ ¼ 2 2 2 2 1 1 0 4 4 m a T m a T Z Z ... (26)

(b) Average potential energy over one complete oscillation.

The instantaneous P.E. of a particle executing SHM is given by 1 2 2sin2 2 U mZ a Zt. 2 2 2 0 0 1 sin 2 T T m a t d t U d t Z Z ? !

³

2 2 0 sin 2 1 4 2 T t m a t T Z Z ª º « » « » ¬ ¼ 2 2 2 2 1 1 0 4 4 m a T m a T Z Z ... (27) (b) Average total energy over one complete cycle.

Since the total energy is independent of time

2 2 1 2 E mZ a ? ! . Example 37

What is the ratio between the maximum to average kinetic energy of a particle executing SHM. Solution

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We known that the maximum K.E., is given by _max 1 2 2 2

K mZ a and average K.E., is given by

2 2 1 4 average K mZ a max 2 1 average K K ? _. Example 38

The average speed over the period of complete oscillation of a particle executing SHM is 2 cm/sec. The particle attains this speed when it is at P, distant 4 cm from the mean position O. Find (i) the amplitude (ii) the time period. Also show that the least time taken by the particle to travel from O to P

is 2 1 2 4 4 sin 4 4 S S §_¨ ·_¸ ¨ ¸

Let a cm and Z rad/sec be the amplitude and angular frequency of oscillation. Then average speed <v> is given by <v> = 2 aZ

S . 2 2 aZ S ? aZ S . ... (1)

Also, the speed (v) at a displacement (x) is given by

2 2 2 2 v Z a x 2 2 2 2 2 v Z a Z x ... (2) 2 2 4 S Z 16 2 2 2 4 4 16 4 S S Z § · Z ¨_¨ ¸_¸ © ¹ 2 4 1 , 4 From a S S cm Z _S ? 2 2 8 sec 4 and From T S S Z _S .

Let the displacement equation for the particle be x = a sin Zt. [Here for the sake of simplicity phase constant I is taken zero].

2 2 4 4 sin 4 4 t S S S .

Let at t = t₁ the particle be at 0

2 1 2 4 4 0 sin 4 4 t S S S ? t1 0

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Now, let the particle be at P, at t = t₂ for the first time Then 2 2 2 4 4 4 sin 4 4 t S S S 2 2 2 4 4 sin 4 t S _S S 2 1 2 2 4 4 sin 4 4 t S S §_¨ ·_¸ ¨ ¸ _© _¹

? The required time interval = t2 – t1

A particle executes SHM whose displacement changes with time as x = a sin Zt. Find the average speed between the instants when it passes its mean position and its displacement is

2

a

for the first time.

Solution

Since the displacement equation is given as x = a sin Ztwhen x = 0; t = 0 and when x = 2

a

; t ₆S_Z. Also, instantaneous velocity v = a ZcosZt.

? Required value <v> = /6 0 /6 0 cos a t d t d t S Z S Z Z Z

³

/6 0 /6 0 sin sin sin 0 6 0 6 t a _a t S Z S Z Z _S Z Z S Z § · _ª _{§ ·} _º ¨ ¸ _« _{¨ ¸} _» ¨ ¸ © ¹ © ¹ ¬ ¼ § · ¨ ¸ © ¹ 1 6 3 2 a u Z Z S S § ·_u ¨ ¸ © ¹ .

General Method to show that a given body (or a particle) executes SHM and to find its time period. Step 1: Establish the equilibrium condition (if any) for the body at its mean position.

Step 2: Imagine the body to be displaced (linear or angular as the case may be) from its mean position slightly. Find the net restoring force (in case of linear SHM) or torque (in case of angular SHM) acting on the body in terms of displacement, ending up finally in the form of

F = – kx (for linear SHM)

or W kT (for angular SHM).

Where k should be a constant.

Step 3 : Dividing F by M or (W by I, as the case may be) find the net acceleration as :

2 2 2 d x x dt Z or 2 2 2 d dt T Z T

Knowing the value of Z, time period can be calculated as 2

T S .

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Alternatively: One can also find the total energy of the system, in step 2 and differentiate the expression with respect to time t and equate it to zero. [ The T.E. in SHM remains constant]. One should then eventually end up in equation of the form as obtained in step 3. [See example 25, alternative]. A few examples are solved stepwise to illustrate the idea.

EXAMPLES Example 1

Write down the equation of a simple harmonic motion with an amplitude of 5 cm if 150 oscillations are performed in one minute and the initial phase is 450_.

Solution

Given: Amplitude (A) = 5 cm Initial phase G 450 S / 4 rad.

Frequency of oscillation = n = 150 per minute = 150 60 per second 5 2 per second ? Angular frequency = 2 2 ( )5 5 1 2 n rad s Z S S S

? The displacement of the particle at any instant t is given by

sin 5sin 5

4

x A ZtG §_¨ StS ·_¸ cm

Example 2

In what value of time t (i.e., after t = 0) will a particle, oscillating according to the equation x = 7 sin 0.5 tS , move from the position of equilibrium to the maximum displacement?

Solution

The displacement of the particle is given by x = 7 sin 0.5 tS Given: x_max = amplitude = 7 Alternate Method:

7 7 sin 0.5 tS ? 2 0.5 T 4 sec T S S 1 sin 0.5 tS ? At t = 0, x = 7 sin (0) = 0 0.5 2 t S S

? Time to reach from x = 0

1 sec

t ond

? i.e., equilibrium) to x = 7

(Amplitude) that is a quarter Oscillation t = 1 sec. Example 3

The equation of an oscillating particle is given by 2 sin ( )

2 4

x S tS cm. Find (i) Period of oscillation (ii) Maximum velocity

(iii) Maximum acceleration and (iv) Initial displacement

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Solution

Given that the displacement of the particle is 2 sin ( )

2 4

x S tS cm

Compare it with x = A sin ZtG

? Amplitude (A) = 2 cm Angular frequency 1 2 rad s S Z .

(i) Period of oscillation 2 2 4 / 2

T S S s

Z S

(ii) Maximum velocity vmax ZA =

1

2

2 cm s

S_u_S

(iii) Maximum acceleration = amax Z2 A =

2 2 2 / 2 2 2 cm s S S _u

(iv) Initial displacement = 0 2 sin ( ) 2

4

x S cm. (i.e., at t = 0).

Example 4

The equation of a simple harmonic motion is given as x 6 sin10St8 cos 10St

Where x is in cm and t is in seconds. Find the amplitude, period and initial phase. Solution

Put x 6sin10St8 cos 10St Asin ZtD

Expanding, we get, x AcosD sinZtAsinDcosZt 1 , 10 We get Z S rad s ? 2 2 0.2 10 The Period T S S s Z S ? _.

Also A cos D= 6 and A sin D= 8

Squaring and adding, we get A2_{= 6}2_{+ 8}2_{= 100}

? Amplitude = A = 10 cm. sin 8 tan 1.333 cos 6 A A D D D ? Initial phase = D= 530_8’. Example 5

A particle performing S. H. M has velocities of 3 cm s-1_{and 4 cm s}-1_{respectively when the displacements}

are 4 cm and 3 cm. Find the amplitude, period and maximum velocity of the particle. Solution

The velocity v of the particle performing SHM when its displacement x is given by

2 2 v Z A x 2 2 2 2 v Z A x ? We have, v = 3 cm s-1_{when x = 4 cm.}

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2 2 9 Z A 16 ? … (I) Also, v = 4 cm s-1_{when x = 3 cm.} ? 2 2 16 Z (A 9) … (II)

Dividing (I) by (II),

2 2 9 16 16 9 A A Solving, A = 5 cm

Substituting this in (I), we get Z= 1 rad s–1

? Period T = 2 /S Z 2 sS Velocity is maximum, when x = 0

1

max 1.0 5

v ZA cm s

? u

Example 6

A particle executing linear SHM had a maximum, velocity of 40 cm s–1_{and a maximum acceleration of}

50 cm s-2_{. Find its amplitude and the period of oscillation.}

Solution

Let A be the amplitude and Zbe its angular velocity..

Then maximum velocity v_max ZA 40cm s1 ... (I) Maximum acceleration a_max Z2 A 50cm s2 ... (II) Dividing (II) by (I), we get,

2 1 50 . ., 1.25 40 A i e rad s Z _Z Z ? Period of oscillation = T = 2 2 5.03 1.25 s S S Z Amplitude = A = 40 32 1.25 A cm Z Z . Example 7

A particle of mass 50 g executes SHM along a path of length 20 cm at the rate of 600 oscillations per minute. Find its total energy. Also find the potential and kinetic energies when the displacement x = A/2 where A is the amplitude.

Solution: mass (m) = 50 g = 5 × 10-2_kg Angular frequency 2 600 20 1 60 rad s S Z u S Amplitude (A) 20 10 0.1 2 cm m ? Total energy = 1 2 2 2 mZ A 2 2 2 1 5 10 400 10 0.987 2 S J u u u u . When, x = 2 A , Potential energy = 2 2 2 2 1 1 0.247 2 2 4 A mZ x mZ J ? Kinetic energy = 1 2 2 2 2mZ A x 2 2 2 1 ( ) 0.740 2 4 A mZ A J.

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Example 8

A body is supported by a light spiral spring and causes a stretch of 1.5 cm in the string. If the mass is now set in vertical oscillation of small amplitude, what is the periodic time of the oscillation? Solution

Let m be the mass of the body in kg.

Then, since 1.5 cm = 0.015 m, mg = k × 0.015, at equilibrium position ... (I) Where k is the spring constant in N m–1

K = 1

0.015

m g N m

If the mass is displaced by x below its equilibrium position, the net restoring force F = K (0.015 + x) – mg = Kx

i.e. the net force (restoring) is opposite to the displacement i.e., F K x

)& &

i.e., the motion is Simple Harmonic

2 m T period K S ? = 0.015 0.015 2 2 9.81 m m g S S _{= 0.25 s}

1. A body of mass 1 kg is executing harmonic motion which is given by x = 6.0 cos 100tS/ 4 cm. What is the (i) amplitude of displacement, (ii) frequency, (iii) initial phase, (iv) velocity, (v) acceleration, (vi) maximum kinetic energy?

Solution:

The given equation of SHM is x = 6.0 cos 100tS/ 4 cm

Comparing it with the standard equation of SHM, x = A cos ZtI , we have (i) Amplitude A = 6.0 m

(ii) Frequency Z = 100/sec. (iii) Initial phase I S/ 4

(iv) Velocity v Z A2x2 100 36x2

(v) Acceleration Z2x 100 2x 104x

(vi) Kinetic energy 1 2 1 2 2 2

2m v 2mZ A x . When x = 0, the kinetic energy is maximum i.e.,

2.

A particle is executing SHM with amplitude A and has maximum velocity V

₀. Its sped at displacement A/2 will be : a. 3 V0/ 2 b. V₀/ 2 c. V₀ d. V₀/4 Solution : (A) 2 2 v Z A x

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For x = A/2 2 2 / 2 v Z A A 0 0 3 3 / 4 [ ] 2 V v ZA V ZA

3. The velocity time graph of a harmonic oscillator is shown in the following figure. The frequency of oscillation is : a.25 Hz b. 50 Hz c. 12.25 Hz d. 33.3 Hz. Solution: (A) T = 0.04 s 1 1 25 . 0.04 v Hz T

4. A body is executing SHM. When its displacement from the mean position is 4 cm and 5 cm, the corresponding velocity of the body is 10 cm per sec. and 8 cm per sec. respectively. Then the time period of the body is

a. 2S sec b. 2 S sec c. S sec d. 3 2 S sec Solution: (C) Using 2 2 v Z A x 2 2 10 Z A 4 ... (i) & 2 8 Z A 5 ... (ii) (i) 2 2 10 2 4 A Z § · _¨ _¸ © ¹ (ii) 2 2 8 2 5 A Z § · _{¨ ¸} © ¹ 2 2 2 2 100 64 4 5 Z Z 2 36 6 9 3 Z Z ? 2 2 6 / 3 T v S S _S ?

5. A linear harmonic oscillator has a total mechanical energy of 200 J. Potential energy of it at mean position is 50 J. Find.

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a. The minimum kinetic energy b. The minimum potential energy

c. The potential energy at extreme positions. Solution:

At mean position, potential energy is minimum and kinetic energy is maximum. Hence U_min = 50 J (at mean position) and K_max = E – U_min = 200 – 50

= 150 J (at mean position). At extreme positions, kinetic energy is zero and potential eneryg is maximum.

max

U E

? = 200 J (at extreme position). Single Answer Correct:

1. A particle is executing SHM. Which of the following is maximum when the bob is at the mean position?

a. Time period b. Amplitude c. velocity d. acceleration.

Solution: (C)

At the mean position, velocity is maximum and acceleration is minimum. 2. The displacement of a particle in simple harmonic motion in one time period is

a.A b. 2 A c. 4 A d. zero.

Solution: (D)

At the end of one complete vibration, the particle returns to the initial position.

3. Two bodies A and B of equal masses are suspended from two separate mass less springs of spring constants k₁ and k₂ respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is:

a.k₁ / k₂ b. k1/k2 c. k2/k1 d. k2/k .1 Solution : (D) 1 1 2 1 1 2 2 2 2 1 a k a a or a k Z Z Z Z

4. The equation of SHM of a particle is a + 4_S2_{x = 0, where ‘a’ is instantaneous linear acceleration at}

displacement x. The frequency of motion is

a.1 Hz b. 4SHz c. 1 4 Hz d. 4 Hz Solution: (A) Comparing with 2 2 2 0, 4 ; 2 2 , 1 . aZ x Z S Z Sv S v H z

5. A particle executes SHM of amplitude 5 cm and period 3 s. The velocity of the particle at a distance 4 cm from the mean position (take S= 3) is

a.8 cm s-1 _{b. 12 cm s}-1 _{c. 4 cm s}-1 _{d. 6 cm s}-1_.

Solution:

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2 2 2 2 2 v a x a x T S Z 2 3 1 25 16 6 3 cms u

6. Figure shows the displacement – time graph of a simple harmonic oscillator. The amplitude, time period and initial phase of the oscillator, are respectively.

a.1 cm, 2 s, S/2 b. 2 cm, 4 s, S/4 c. 2 cm, 4 s, zero d. 2 cm, 2 s, zero. Solution: (C)

Graph begins from the origin. So, initial phase is zero. Multiple Answers:

12. A particle vibrates in SHM along a straight line. Its greatest acceleration is 5S2_{cm s}-2_{and when its}

distance from the equilibrium position is 4 cm, the velocity of the particle is 3S cm s-1_{. Then:}

a. The amplitude i 10 cm b. The period of oscillation 2 sec. c. The amplitude is 5 cm d. The period of oscillation 2 sec. Solution: (B), (C) (Acceleration)_max = 2 2 5 a Z S . ... (1) Velocity at y = 4 cm is 2 2 2 2 4 3 v Z a y Z a S or, 2 16 3 a Z S Squaring, Z2 a216 9S2 ... (2) So, 2 2 ₂ 2 2 16 ₉ ₉ 5 5 a a Z _S Z S or, 2 16 9 9 41 5 5 10 a or a a r u or - 3.2 cm. ? a = 5 cm. From (1), Z2u 5 5S2. or, Z S . ? T = 2 sec.

So, choice (B) and (C) are correct and choice (A) and (D) are wrong.

13. When the particle is half way to its end point in a S. H. M (where E is the total energy) : a. 4 E PE b. 5 4 E PE c. PE = KE d. KE = 3 PE. Solution : (A), (D) Given, 2 a

y and total energy = E

We also know that total of energy a particle in SHM 1 2 2 2

E mZ a

Potential energy of particle at some point (PE)

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Choice (D) is correct and choices (B) and (C) are wrong. EXERCISE Single Answer Correct.

22. If the displacement equation of a particle be represented by y = A sin pt + B cos pt, the particle executes.

a. a uniform circular motion b. a uniform elliptical motion.

c. a SHM d. a rectilinear motion.

26. What fraction of the total energy is potential when the displacement is one - half of the amplitude? a. 1 4 b. 2 4 c. 3 4 d. 3.5 4 . 27. The displacement of a particle executing simple harmonic motion is given by : y = 10 sin 6

3

t S

§ ·

¨ ¸

Here, y is in meter and t is in second. The initial displacement and velocity of the particle are respectively. a. 5 3 and 30 m s-1

b. 20 3 and 30 m s-1_.

c. 15 3 m and 30 m s-1

d. 15 m and 5 3 m s-1_.

28. The equation of a SHM of amplitude ‘a’ and angular frequency ' 'Z in which all distances are measured from one extreme position and time is taken to be zero at the other extreme position is

a. x = a sin Zt b. x = a cos Zt

c. x = a – a sin Zt d. x = a + a cos Zt.

29. The amplitude and the periodic time of SHM are 5 cm and 6 s respectively. At a distance of 2.5 cm away from the mean position, the phase will be:

a. S/6 b. S/4 c. 5S/12 d. S/3.

30. Two particles execute SHM of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions, each time their displacement is half of the their amplitude. The phase difference between them is:

a. 900 _{b. 30}0 _{c. 120}0 _{d. 60}0_.

31. The maximum displacement of a particle executing SHM is 1 cm and the maximum acceleration is (1.57)2_{cm s}-2_{. Then the time period is}

a. 0.25 s b. 4.00 s c. 1.57 s d. (1.57)2_s.

32. Which of the following quantities are always positive in a simple harmonic motion? [Letters have usual meanings].

a. .F a )& & b. .v r & & c. .a r & & d. .F r_s )& )& .

33. The ratio of the KE and PE possessed by a body executing SHM when it is at a distance of 1/n of its amplitude from the mean position is

1