Boundary Conditions

The two solutions to the general SHM equation are and . Which solution applies to a particular system depends, as mentioned above, on the boundary conditions for that system. For systems such as the harmonic oscillator and the simple pendulum, the boundary condition that gives the solution

is that the displacement x = x₀ when t = 0. For some other systems it might turn out that x = 0 when t =0. This will lead to the solution . From a practical point of view, the two solutions are essentially the same; for example when timing the oscillations of a simple pendulum, you might decide to start the timing when the pendulum bob passes through the equilibrium position. In effect, the two solutions differ in phase by .

The table in Figure 411 summarises the solutions we have for SHM. x = x ₀ cosωt x = x ₀ sinωt ν = − ν ₀ sinωt ν = ν ₀ cosωt ν = – ωx ₀ sinωt ν = – ωx ₀ cosωt ν = ± ω

_√

______ x₀2 – x2 ν = ± ω

_√

______ x₀2 – x2

Figure 411 Common equations

We should mention that since the general solution to the SHM equation is there are in fact three solutions to the equation. This demonstrates a fundamental property of second order differential equations; that one of the solutions to the equation is the sum of all the other solutions. This is the mathematical basis of the so-called principle of superposition.

4.1.6 SOLVING SHM PROBLEMS

In this section we look at an example of a typical SHM problem and its solution.

Examples

The graph in Figure 412 shows the variation with time t of the displacement x of a system executing SHM.

-10 -8 -6 -4 -2 0 2 4 6 8 10 0. 5 1 1. 5 2 2. 5 3 3. 5 t /s x /c m

Figure 412 Displacement – time graph for SHM

Use the graph to determine the

(i) period of oscillation (ii) amplitude of oscillation (iii) maximum speed (iv) speed at t = 1.3 s

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Solutions

(a) (i) 2.0 s (time for one cycle) (ii) 8.0 cm

(iii) Using gives

(remember that ) = 25 cm s-1

(iv) v = −v₀sinωt = −25sin (1.3π). To find the value of the sine function, we have to convert the 1.3π into degrees (remember ω and hence

ωt, is measured in radians)

1 deg therefore 1.3π = 1.3 × 180

= 234°

therefore v₁ = −25sin (234°) = +20 cm s –1_.

Or we can solve using ω

_√

_______

(

x₀2 _{– x}2

₎

from the graph at t = 1.3 s, x = – 4.8 cm

therefore v = π × = 20 cm s-1

(v) Using = π2 _{× 8.0 = 79 m s}-2

Exercise 4.1

1. (a) Answer the same questions (a)(i) to (a)(iv) in the above example for the system oscillating with SHM as described by the graph in the Figure below.

(b) Also state two values of t for when the magnitude of the velocity is a maximum and two values of t for when the magnitude of the acceleration is a maximum. -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 0. 5 1 1. 5 2 2. 5 3 3. 5 t /s x/ cm

4.2 ENERGY

CHANGES

DURING SIMPLE

HARMONIC

MOTION SHM

4.2.1 Describe the interchange between kinetic energy and potential energy during SHM

4.2.2 Apply the expression EK = ½mω

2_(x

0

2 _{– x}2_{) for} the kinetic energy of a particle undergoing simple harmonic motion, ET = ½mω2x0

2 _{for the} total energy and EP = ½mω

2_x2 _{for the potential} energy.

4.2.3 Solve problems, both graphically and by calculation, involving energy changes during simple harmonic motion.

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4.2.1 KINETICANDPOTENTIAL

ENERGYCHANGES

We must now look at the energy changes involved in SHM. To do so, we will again concentrate on the harmonic oscillator. The mass is stationary at x = +x₀ (maximum extension) and also at x = –x₀ (maximum compression). At these two positions the energy of the system is all potential energy and is in fact the elastic potential energy stored in the spring. This is the total energy of the system

E_T and clearly

Equation 4.12

That is, that for any system performing SHM, the energy of the system is proportional to the square of the amplitude. This is an important result and one that we shall return to when we discuss wave motion.

At x = 0 the spring is at its equilibrium extension and the magnitude of the velocity v of the oscillating mass is a maximum v₀. The energy is all kinetic and again is equal to

E_T. We can see that this is indeed the case as the expression for the maximum kinetic energy E_max in terms of v₀ is

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Clearly E_T and E_max are equal such that

E_T = 1 __ 2 kx0 2 _{= E} max = 1 __ ₂ mv0 2 From which v₀2 _{= k __} m x0 2 _{(as v} 0 ≥ 0) Therefore v₀ =

√

__ k __ _{m x0} = ωx₀

which ties in with the velocity being equal to the gradient of the displacement-time graph (see 4.1.5). As the system oscillates there is a continual interchange between kinetic energy and potential energy such that the loss in kinetic energy equals the gain in potential energy and E_T = E_K + E_P.

4.2.2 THE SHM ENERGYEQUATIONS

Remembering that , we have that

Equation 4.14

Clearly, the potential energy E_P at any displacement x is given by

Equation 4.15

At any displacement x, the kinetic energy E_K is E_K= 1__ 2mv2 Hence remembering that

v _______

(

x₀2 _{– x}2

₎

_{, we have} E_K= 1__₂m 2

(

x₀2 _{– x}2

)

Equation 4.16

Although we have derived these equations for a harmonic oscillator, they are valid for any system oscillating with SHM. The sketch graph in Figure 414 shows the variation with displacement x of E_K and E_P for one period.

ener

gy

potential kinetic

Figure 414 Energy and displacement

4.2.3 SOLVING SHM ENERGY

PROBLEMS

Let us now look at solving a problem involving energy in SHM.

Example

The amplitude of oscillation of a mass suspended by a vertical spring is 8.0 cm. The spring constant of the spring is 74 N m–1_{. Determine}

(a) the total energy of the oscillator

(b) the potential and the kinetic energy of the oscillator at a displacement of 4.8 cm from equilibrium.

Solution

(a) Spring constant k = 74 N m-1 _and

x₀ = 8.0 × 10-2 _m E_T = ½ kx₀2 = ½ × 74 × 64 × 10-4 _{= 0.24 J} (b) At x = 4.8 cm Therefore E_P = ½ × 74 × (4.8 × 10-2₎2 = 0.085 × 10-4 _J E_K = E_T − E_P = 0.24 − 0.085 = 0.16 J Exercise 4.2

In a simple atomic model of a solid, the atoms vibrate with a frequency of 2.0 × 1011 _{Hz. The amplitude of vibration}

of the atoms is 5.5 × 10–10 _{m and the mass of each atom is}

4.8 × 10–26 _{kg. Calculate the total energy of the oscillations}

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4.3.1 State what is meant by damping.

4.3.2 Describe examples of damped oscillations.

4.3.3 State what is meant by natural frequency of

vibration and forced oscillations.

4.3.4 Describe graphically the variation with forced frequency of the amplitude of vibration of an object close to its natural frequency of vibration.

4.3.5 State what is meant by resonance.

4.3.6 Describe examples of resonance where the effect is useful and where it should be avoided.

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4.3.1 DAMPING

In this section, we look at oscillations of real systems. In section 4.1.3, we described an arrangement by which the oscillations of a pendulum could be transcribed onto paper. Refer to Figure 415.

time di s p la c e ment Figure 415 Damping

The amplitude of the oscillations gradually decreases with time, whereas for SHM, the amplitude stays at the same value forever. Clearly, the pendulum is losing energy as it oscillates. The reason for this is that dissipative forces are acting that oppose the motion of the pendulum. As mentioned earlier, these forces arise from air resistance and though friction at the support. Oscillations, for which the amplitude decreases with time, are called damped

oscillations.

4.3.2 EXAMPLES

OFDAMPED

OSCILLATIONS

All oscillating systems are subject to damping as it is impossible to completely remove friction. Because of this, oscillating systems are often classified by the degree of damping. The oscillations shown in Figure 415 are said to be lightly damped. The decay in amplitude is relatively slow and the pendulum will make quite a few oscillations before finally coming to rest. Whereas the amplitude of the oscillations shown in Fig 416 decay very rapidly and the system quickly comes to rest. Such oscillations are said to be heavily damped. time am p li tude

Figure 416 Heavily damped oscillations

Consider a harmonic oscillator in which the mass is pulled down and when released, and the mass comes to rest at its equilibrium position without oscillating. The friction forces acting are such that they prevent oscillations. However, suppose a very small reduction in the friction forces would result in heavily damped oscillation of the oscillator, then the oscillator is said to be critically

damped.

The graph in Figure 417 shows this special case of damping known as critical damping.

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0 0.2 0.4 0.6 0.8 1 1.2 0 0.2 0.4 0.6 0.8 1 1.2 time / s d is p la ce m e n t / m

Figure 417 Critical damping

Although not in the IB syllabus, a useful way of classifying oscillating systems, is by a quantity known as the quality

factor or Q-factor. The Q-factor does have a formal

definition but it is approximately equal in value to the number of oscillations that occur before all the energy of the oscillator is dissipated. For example, a simple pendulum has a Q-factor of about 1000.

As mentioned in the introduction to this chapter, the oscillations (vibrations) made by certain oscillatory systems can produce undesirable and sometimes, dangerous effects. Critical damping plays an important role in these situations. For example, when a ball strikes the strings of a tennis racquet, it sets the racquet vibrating and these vibrations will cause the player to lose some control over his or her shot. For this reason, some players fix a “damper” to the springs. If placed on the strings in the correct position, this has the effect of producing critically damped oscillations and as a result the struck tennis racquet moves smoothly back to equilibrium. The same effect can be achieved by making sure that the ball strikes the strings at a point known as the ‘sweet spot’ of which there are two, one of which is know as the ‘centre of percussion (COP)’. Cricket and baseball bats likewise have two sweet spots.

Another example is one that involves vibrations that may be set up in buildings when there is an earthquake. For this reason, in regions prone to earthquakes, the foundations of some buildings are fitted with damping mechanisms. These mechanisms insure any oscillations set up in the building are critically damped.

Exercise 4.3

Identify which of the following oscillatory systems are likely to be lightly damped and which are likely to be heavily damped.

1. Atoms in a solid 2. Car suspension 3. Guitar string

4. Harmonic oscillator under water 5. Quartz crystal

6. A cantilever that is not firmly clamped 7. Oil in a U-tube

8. Water in a U-tube

4.3.3 NATURAL

FREQUENCYAND

FORCED

OSCILLATIONS

Consider a small child sitting on a swing. If you give the swing a single push, the swing will oscillate. With no further pushes, that is energy input, the oscillations of the swing will die out and the swing will eventually come to rest. This is an example of damped harmonic motion. The frequency of oscillation of the swing under these conditions is called the natural frequency of oscillation (vibration). So far in this topic, all the systems we have looked at have been systems oscillating at their natural frequency.

Suppose now when each time the swing returns to you, you give it another push. See Figure 418

Figure 418 A forced oscillation

The amplitude of the swing will get larger and larger and if you are not careful your little brother or sister, or who ever the small child might be, will end up looping the loop.

The frequency with which you push the swing is exactly equal to the natural frequency of oscillation of the swing and importantly, is also in phase with the oscillations of the spring. Since you are actually forcing the swing to oscillate,

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the swing is said to be undergoing forced oscillations. In this situation the frequency of the so-called driver (in this case, you) is equal to the natural frequency of oscillation of the system that is being driven (in this case, the swing). If you just push the swing occasionally when it returns to you, then the swing is being forced at a different frequency to its natural frequency. In general, the variation of the amplitude of the oscillations of a driven system with time will depend on the

• frequency of the driving force • frequency of natural oscillations • amplitude of the driving force

• phase difference between driving force frequency and natural frequency

• amount of damping on the system

(There are many very good computer simulations available that enable you to explore the relation between forced and natural oscillations in detail.)

The driving force and system are in phase if, when the amplitude of system is a maximum, it receives maximum energy input from the driver. Clearly this is when the amplitude of the driver is a maximum.

What is of particular interest is when the forced frequency is close to and when it equals the natural frequency. This we look at in the next two sections.

4.3.4 FORCED

FREQUENCYAND

AMPLITUDE

We now look at how the amplitude of an oscillating system varies with the frequency of the driving force.

The graph in Figure 419 shows the variation with frequency

f of the driving force of the amplitude A of three different

systems to which the force is applied.

0 5 10 0 5 10 15 20 25 30 f/ Hz A /cm heavy damping medium damping light damping

Figure 419 Forced frequency

Each system has the same frequency of natural oscillation,

f₀ = 15 Hz. The thing that is different about the systems is that they each have a different degree of damping: heavy (low Q), medium (medium Q), light (large Q).

For the heavily damped system we see that the amplitude stays very small but starts to increase as the frequency approaches f₀ and reaches a maximum at f = f₀; it then starts to fall away again with increasing frequency.

For the medium damped system, we see that as f approaches

f₀, the amplitude again starts to increase but at a greater rate than for the heavily damped system. The amplitude is again a maximum at f = f₀ and is greater than that of the maximum of the heavily damped system.

For the lightly damped system, again the amplitude starts to increase as f approaches f₀, but at a very much greater rate than for the other two systems; the maximum value is also considerably larger and much more well-defined i.e. it is much easier to see that the maximum value is in fact at f = f₀.

If there were such a thing as a system that performs SHM, then if this system were driven at a frequency equal to its natural frequency, its amplitude would be infinite. Figure 420 shows how the amplitude A for a driven system with very little damping and whose natural frequency of oscillation f₀ = 15 Hz , varies with the frequency f of the driving force. 0 50 100 150 0 5 10 15 20 25 f /Hz 30 A /cm

Figure 420 Amplitude-frequency graph for a lightly damped oscillator

We see that the maximum amplitude is now very large and also very sharply defined. Also, either side of f₀, the amplitude drops off very rapidly.

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4.3.5 RESONANCE

(Note: As well as the availability of a large number of computer simulations that demonstrate resonance, there are also many laboratory demonstrations and experiments that can be done to demonstrate it).

We have seen that when an oscillatory system is driven at a frequency equal to its natural frequency, the amplitude of oscillation is a maximum. This phenomenon is known as resonance. The frequency at which resonance occurs is often referred to as the resonant frequency.

4.3.6 EXAMPLESOFRESONANCE

In the introduction to this Topic, we referred to resonance phenomena without actually mentioning the term resonance. For example we can now understand why oscillations in machinery can be destructive If a piece of machinery has a natural frequency of oscillation and moving parts in the machine act as a driver of forced oscillations and have a frequency of oscillation equal to the natural frequency of oscillation of the machine, then the amplitude of vibration set up in the machine could be sufficient to cause damage.

Similarly, if a car is driven along a bumpy road, it is possible that the frequency with which the bumps are crossed by the car, will just equal the natural frequency of oscillation of the chassis of the car, If this is the case then the result can be very uncomfortable.

We now also see why it is important that systems such as machines, car suspensions, suspension bridges and tall buildings are critically or heavily damped.

Resonance can also be very useful. For example, the current in a particular type of electrical circuit oscillates. However, the oscillating current quickly dies out because of resistance in the circuit. Such circuits have a resonant frequency and if driven by an alternating current supply, the amplitude of the current may become very large, particularly if the resistance of the circuit is small. Circuits such as this are referred to as resonant circuits. Television and radios have resonant circuits that can be tuned to oscillate electrically at different frequencies. In this way they can respond to the different frequencies of electromagnetic waves that are sent by the transmitting station as these waves now act as the driving frequency. This is discussed in more detail in Topic F.1

In the introduction we mentioned the use of quartz crystal as timing devices. If a crystal is set oscillating at its natural frequency, electric charge constantly builds up and dies away on it surface in time with the vibration of the crystal (This is known as the piezoelectric effect.). This makes it easy to maintain the oscillations using an alternating voltage supply as the driving frequency. The vibrations of the crystal are then used to maintain the frequency of oscillation in a resonant circuit. It is the oscillations in the resonant circuit that control the hands of an analogue watch or the display of a digital watch. (The concept of analogue and digital signals is discussed in Topics 14.1 and C.1).

It is left as an exercise to you to think of other situations in which resonance can be useful or can be harmful.

4.4 WAVE

CHARACTERISTICS

4.4.1 Describe a wave pulse and a continuous progressive (travelling) wave. (Students

should be able to distinguish between oscillations and wave motion and appreciate that in many examples, the oscillations of the particles are simple harmonic).

4.4.2 State that progressive (travelling) waves transfer energy. (Students should understand that there is no net motion of the medium through which the wave travels).

4.4.3 Describe and give examples of transverse and of longitudinal waves.

4.4.1 Wave pulses and continuous travelling waves

© IBO 2007

Perhaps one of the most familiar types of wave motion is a water wave. However, we can also set up waves in strings very easily. A simple demonstration is to take a length of rubber tubing. Hold one end of it and shake that end up and down. A wave will travel down the tube. If we give the end of the tube just one shake then we observe a pulse to travel down the tube. By this we can see that we can have either a continuous travelling wave or a travelling pulse. This is illustrated in Figure 421 in which we have taken an instantaneous snap shot of the tube.

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pulse hand movement (a) (b)

Figure 421 Simple waves

We can also set up another type of wave by using a slinky spring. In this demonstration we lay the spring along the floor. Hold one end of it and move our hand backwards and forwards in the direction of the spring. In this way we see a wave travelling down the spring as a series of compressions and expansions of the spring as illustrated in Figure 422.

We can also set up a pulse in the spring by moving our hand backwards and forwards just once in the direction of the spring.

Of course we can set up a wave in the spring that is similar to the one we set up in the rubber tube. We shake the spring in a direction that is at right angles to the spring as shown in Figure 422.

compressions expansions

compress expand

compress expand compress

2(a) 2(b)

Figure 422 Slinky springs

A very important property associated with all waves is their so-called periodicity. Waves in fact are periodic both in time and space and this sometimes makes it

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