A motion which repeat itself after a fixed interval of time is called periodic motion. Example : Motion of Earth around the sun etc.

O S C I L L A T IO N S

PERIODIC MOTION :

A motion which repeat itself after a fixed interval of time is called periodic motion.

Example : Motion of Earth around the sun etc.

OSCILLATORY MOTION :

Its a type of periodic motion which performs its motion in terms of to and fro about a point such as motion of a perfectly elastic ball on perfectly rigid floor (dropping a ball up & down).

HARMONIC MOTION :

Its a type of oscillatory motion which can be represented by sine or cosine function.

Example : Variation of Kinetic and Potential Energy of simple pendulum or SHM of spring mass system,

simple pendulum etc.

SIMPLE HARMONIC MOTION (S.H.M.) :

It is simplest form of Harmonic Motion in which force or acceleration acting on it is directly proportional to its displacement and its direction is always opposite to displacement.

Example : Simple pendulum, spring mass system etc. Periodic Motion

Oscillatory Motion Harmonic Motion Simple Harmonic Motion

Few Important Points

(i) All S.H.M. are Harmonic motion and all harmonic motion are Oscillatory and all Oscillatory are Periodic but reverse is not necessary.

(ii) All examples given is specific examples of the heading.

As per the definition F  –x F = –kx where k is constant ma = –kx 2 2 dv d x a dt dt   2 2 d x k x 0 dt m  ...(1)

It is a differential equation of S.H.M. which can have many equations (solutions).

2 2 d x k x 0 dt m  (a) Solution-1 x = A sin or x = A sin (b) Solution-2 x = A cos or x = A cos (c) Solution-3 x = a cos or x = a sin + bcos + bcos

www.justphysics.in Where 2 k

x

  & A &  = constant Let us Varify it : (a) x = A sin t diff. w.r.t. ‘t’ dx A cos t dt    again diff. w.r.t. ‘t’ 2 2 2 d x A cos t dt     or = –²x or 2 2 2 d x x 0 dt    standard eqn of SHM same we can prove for x = Asin(t + )

(b) x = A cos t dx A sin t dt     , 2 2 2 2 2 2 d x d x A cos t x 0 dt      dt    (c) x = a sin t + b cos t dx a cos t b sin t dt        2 2 2 2 d x a cos t b cos t dt          2

_

a sin t b cos t

_

2 x   2 2 2 d x x 0 dt     NOTE :

(i) x = Asin t, is a standard equation of S.H.M., where t = 0, x = 0 i.e. body starts from mean position.

(ii) x = A cos t, is a standard equation of S.H.M., where t = 0, x = A i.e. body starts fromnean position.

(iii) x = A sin (t + ), x = a sin t + b cos t are the standard equation of S.H.M., when body start some where other than mean position or extreme position.

DISPLACEMENT, VELOCITY AND ACCELERTION

OF A BODY PERFORMING S.H.M.

Case-1 :

As a function of time ‘t’

(i)

Displacement :

x = A sin t x A –A T t 0 T 2 T=

(ii)

Velocity :

dx d v dt dt   (A sin t) v T t 0 T 2

vA cos t  , vmax = A

v A sin t 2      _  _   Velocity function is 2 

ahead with displacement function.

(iii)

Acceleration :





dv d a A cos t dt dt     a T t 0 T 2 A –A = –A²sin t





2 aA sin   t a_max = ²A

Acceleration function is  ahead with displacement and 2 

ahead with velocity.

Case-2 :

As a function of position ‘x’ (Displacement)

(i)

Velocity :

v = A cos t 2 A 1 sin t     2 2 x A 1 A    x = A sin t 2 2 v  A x

at x = 0, i.e. mean position

v = ± A |v| =

at x = 0, i.e. mean position

v = 0 A= maximum speed

www.justphysics.in now since v2  2



A2x2



O –A A x v 2 2 2 2 v A x    





2 2 2 2 x v 1 A A      Ellipse

(ii)

Acceleration :

a = –²A sin t

x = 0, at mean position

a = 0 |a| =

x = ±A, at mean position

a = ±

A= maximum

x a

KINETIC ENERGY (KE), POTENTIAL ENERGY (U) AND

MECHANICAL ENERGY (E) OF A BODY PERFORMING S.H.M. :

Case-1 :

As a function of time (t) :

(a)

Kinetic Energy (K) :





2 2 1 1 K mv m A cos t 2 2     v T t 0 T 4 T 2 3T 4 1 2 A² 2 2 2 1 K m A cos t 2    2 2 max 1 K m A 2   Time period = T 2 Frequency = 2 2 2 T     

(If a body forming SHM with frequency f and time period T than its Kinetic Energy will oscillate with frequency 2f and time period T/2)

(b)

Potential Energy (U) :

U = –W_C (–ve work done by conservative force) U Fdx   

_

 

_

madx U T t 0 T 4 T 2 3T 4 1 2 A² 2T 5T 4  m (

_

2)xdx 1m 2x2 2  

2 2 2 1 U m A sin t 2    Time period = T 2 2 2 max 1 U m A 2   Frequency =  

(If a body forming SHM with frequency f and time period T than its Potential Energy will oscillate with frequency 2f and time period T/2)

(c)

Mechanical Energy (E) :

E = K + U 1m 2A2 cos2 t sin2 t 2     _    _ 1m 2A2 2   = constant E T t 0 T 4 T 2 3T 4 1 2 A² _K U E = K+U = 1 2 A² NOTE :

(i) Time period of KE & PE is half as compared to oscillating body hence frequency is double. (ii) Mechanical Energy (E) of oscillating body is constant.

(iii) When KE is maximum then PE is minimum and vice versa.

Case-2 :

As a function of Position-x

(a)

Kinetic Energy (K) :

2 2 2 2 1 1 K mv m A x 2 2        

at x = 0, (Mean position) at x = ±A, (Extreme position) U = 0 2 2 2 1 K m A x 2        x K O x=–A x=A 1 2 K = K =max 1 2

(b)

Potential Energy (U) :

at x = 0, (Mean position)

U = 0

at x = ±A, (Extreme position)

U = 1 2

= maximum _x

U

O

[As per Case-1]

2 2

1

U m x

2

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(c)

Mechanical Energy (E) :

E = K + U 1m 2



A2 x2



1m 2x2 2 2      E O E=K+U = 1 2 A 2 –A 2 –A A 1m 2A2 2   = constant

From the above we can also conclude

2 2 1 E K U m A 2     when K = U max 2 2 1 K 0 m A 2     1m 2



A2 x2



1m 2x2 2   2  max 2 2 1 0 U m A 2     x A 2  

When Kinetic Energy is maximum then Potential Energy is zero and when Potential Energy is maximum then Kinetic Energy is zero.

NOTE :

(i) When K = K_max, U = 0 and when U = U_max, K = 0. (ii) E = constant = K_max = U_max.

(iii) At x A 2

  , K = 0.

Example-1 :

Maximum velocity of a oscillating body (SHM) is v₀. Then find its velocity when body is at half of its maxi-mum value ? [ = angular frequency of the body].

Solution :

U_max = wA = v_o 2 2 2 2 0 2 2 x x v A x A 1 v 1 A A         ...(1) at x A 2 





2 0 2 0 0 A / 2 1 3 v v 1 v 1 v A 4 2      Ans.

Example-2 :

Maximum displacement of an oscillating body performing S.H.M. is A (amplitude). Find its displacement when its velocity is half of its maximum value ?

Solution :

2 2 x v A 1 A

   (As per example-1)

2 0 2 x v v 1 A  

at 2 2 0 0 0 2 2 v v x 1 x v v 1 1 2 2 A 4 A        2 2 x 1 3 1 A 4 4     3 x A 2    Ans.

TIME PERIOD OF A BODY PERFORMING S.H.M. :

(a) Force Method (for Linear SHM)

Time Period of a Body Performing S.H.M.

Method-1

or

(b) Torque Method (for Angular SHM)

Energy Method (For both linear &

Angular SHM) Method-2

METHOD-1

(a) Force Method :

There are following steps to solve questions for getting time period.

Step-1 : Get equilibrium position.

Step-2 : Displace the body through very small displacement ‘x’ from its equilibrium position. Step-3 : Calculate restoring force (force responsible to get the body back). This will be F  –x Solve for few more steps

F_r = –kx where k = any constant ma = –kx F_r = F_net = ma 2 2 d x k x 0 dt m  2 2 d x a dt 

Step-4 : Compare the above differential equation with standard equation of SHM.

i.e. 2 2 2 d x x 0 dt    and 2 T   so 2 k m   k m    and T 2 2 m k      (b) Torque Method :

Step-1 : Get equilibrium position.

Step-2 : Displace the body through small angular displacement ‘’.

Step-3 : Get restoring torque (_r), which will be

r

  



sin, cos1, tan



Solve for few more steps

r k

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I k

    (I = MOI of the body about oscillating point)

2 2 2 2 d k d dt I dt        2 2 d k 0 dt I     

Step-4 : Compare it with standard equation of angular SHM 2 2 2 d 0 dt      2 T    So, 2 k I   , T2 

METHOD-2 : ENERGY METHOD :

Step-1 : Get instantaneous position (other than mean or extreme position. Step-2 : Calculate K.E. and P.E. at that position.

Step-3 : Since Mechanical Energy

E = K + U = constant

Step-4 : Since E = constant so, differentiate it w.r.t. ‘t’ and put it zero.

i.e. dE 0 dt 

Solve further and compare it with standard equation of SHM i.e.

2 2 2 d x x 0 dt    and 2 2 2 d 0 dt     

EXAMPLES BASED ON METHOD-1 AND METHOD-2

Example-1 :

A cylinder is floating on liquid as shown in figure. It density of cylinder is , its length is  , and density of liquid  ( > ) then, find its time period of small displacement.

Solution :

Lets displace the body through small displacement ‘x’. Since cylinder is bal-anced initially i.e. Buoyant force due to dipped cylinder is equal to its weight. Now if we dip it further by displacement ‘x’ then dipped volume will increase and hence Buoyant force will increase. Restoring force will be equal to Buoyant force due to extra submerged (dipped) part.

F_r = (–F_b)_x = –(Ax)g

Let A = area of cross-section of cylinder ma = –(Ag)x





2 2 d x A g Agx dt     2 2 d x g x 0 dt   _{ }    

Compare it with standard equation of SHM,  2 g  2 T 2 g        

Example-2

Find out time period of simple pendulum ?

Solution :

Its a case of rotational SHM

Displace the body by small angular displacement ‘’

mg sin mg _{mg cos}



_{I = m} 0  2 O Restoring torque _r = –(mg sin )  I₀ = –mg , sin =  2 2 2 d m mg dt       2 2 d g 0 dt     

compare it with standard equation

2 2 2 d 0 dt      2 g g         and T 2 g 

Example-3

Find time period of a block attached with spring and placed on smooth surface as shown in figure.

m k

Solution :

Method-1 (Force concept)

Displace it by small displacement ‘x’ F r = –kx  ma = –kx 2 2 d x k x 0 dt m    m m k x=0 x kx

www.justphysics.in 2 k m    and T 2 m k   Method-2 m x=0, v=0

Get instantaneous position

v x m 2 2 1 1 KE, K mv , PE U kx 2 2    and E = K + U = constant 2 2 1 1 mv kx C 2 2  Diff. w.r.t. ‘t’ 2 2 1 dv 1 dx m k 0 2 dt 2 dt  2 dv dx md v k2x 0 dt dt    2 2 d x mv kxv 0 dt    2 2 d x k x 0 dt m   

Compare it with standard equation, k m

  and T 2 m k   Does time period in spring mass system depends on gravity ?

 Lets find three situation

m k m x k kx0 mg equilibrium position mg k(x+x )0 x m kx₀ = mg ...(1) Consider third diagram

F_r = –[k(x + x₀) –mg] = –kx 2 2 2 2 d x d x k m kx x 0 dt    dt m  m T 2 k   

QUESTIONS BASED ON SPRING MASS SYSTEM :

As per the previous example Time period T 2 m

k   . Same time period will also be there is following cases.

(a) m k An inclined surface (b) m Inside tunnel dugged in Earth (c) m Inside lift moving

up or down or (d) m At any planet or at moon k

COMBINATION OF SPRING :

In combination of more than one spring then we have to calculate equivalent spring constant and result is

eq

m T 2

k

  _{, where k}

eq = equivalent spring constant.

(a)

Series Combination :

m k1 k2 k3 k1 k2 k3 x1 x2 x3 m x m keq m x mg keqx k1x1 k1x1 k2x2 k2x2 k3x3 k3x3 m mg k3x3 mg = k x = k x = k x1 1 2 2 3 3 x1 = mg/k1 x2 = mg/k2 x3 = mg/k3 keqx keqx keqx m mg keqx mg = k xeq x =mg keq

www.justphysics.in 1 2 3 xx x x  eq 1 2 3 mg mg mg mg k  k  k  k _{eq 1 2 3} 1 1 1 1 k k k k    

(b)

Parallel Combination :

m keq m k1 k2 k3 mg m mg keqx k1x m k2x k3x mg x

=

By the free body diagram

k₁x + k₂x + k₃x = mg ...(1) and k

eqx = mg ...(2)

From (1) & (2)

k_eq = k₁ + k₂ + k₃

Example-1 :

Find out k

eq in the following cases ?

(a) m k k (b) k k k m 3k 3k 2 3k 2 (c) m 2k k k k

Solution :

(a) m k k m k kx kx k x kx mg kx mg kx kx k = 2k

(as per the parallel combination) eq

(b) Its a case of parallel & series combination k k k m 3k 3k 2 3k 2 m 3k 3k 3k eq 1 1 1 1 3 k 3k 3k3k3k k = keq

(c) Again its a combination of series and parallel combination

m 2k k k k m k 2k 2k m k k k = k + k = 2keq

Example-2 :

Find out time period of small oscillation in the following cases ?

(a) k m (b) k m (c) k m

Solution :

(a) It is simple case

k m x

k

m x

www.justphysics.in

If we pull block by distance ‘x’ then spring also expands by x so, T 2 m

k  

(b) [As per the constraint relation given in laws of motion, its a case no.-2] Restoring force F_r = T ...(1) and 2T kx 2  x/2 m T T x 2T k T x 4  ...(2) from (1) & (2) r k F x 4   2 2 d x k m x dt 4    2 2 d x k x 0 dt 4m   

Comparing with standard equation

2 k k 4m 4m      and T2  4m T 2 k  

(c) Again its a second case of constraint equation. F r = –T ...(1) and T k2x 2  k m T x T/2 T/2 2x T 4kx   ...(2) from (1) & (2) F_r = –4kx 2 2 d x m 4kx dt   2 2 d x 4k x 0 dt m   _{ }    2 4k 2 m , T 2 m 4k       

Example-3 :

In this given diagram, find out time period

k k

x x1 x1

of hanging mass for small oscillations ?

Solution :

k k x x1 x1

Displace the block by ‘x’, let spring displaces by x₁ in that case so, x cos  = x₁ ...(1)

FBD :

m kx1 kx1

Restoring force in this case is





r 1 1 F   kx cos kx cos = –2kx₁cos  F_r = –2kx cos² F r = –2k cos²x 2 2 2 d x m 2k cos x dt     2 2 2 d x 2k cos x 0 dt m    

Compare it with standard equation

2 2 2 d x x 0 dt    2 2 2k cos m , T 2 m 2k cos      

Example-3(a)

k k m

If inthe above case we put  = 00 _{then this question will come cos0}0 _{= 1}

m T 2

2k  

PHYSICAL PENDULUM :

Any physical structure which can angularly oscillates about a fixed point is called physical pendulum. Restoring torque   r



mg sin



d

www.justphysics.in





0 I  mgd sin 2 2 0 d mgd 0 dt I     _{ }    mg cos mg mg sin C.M. d O 2 2 2 d 0 dt       2 0 mgd I   0 I 2 T 2 mgd     

Where I₀ = M.O.I. about oscillating point

d = Distance of C.O.M. from oscillating point.

Example :

Find out time period of small oscillation of given structures ?

(a) m O (b) O Uniform rod of mass ‘m’ length (c) O Uniform ring of mass ‘m’ and radius R (d) O Uniform disc of mass ‘m’ and radius R

Solution :

In all cases we will use _{T 2} I0

mgd   Where I

0 = M.O.I about oscillating point, and d is distance of C.O.M. from oscillating body.

(a) Simple pendulum d = _{ ,} I₀m2 (b) Uniform rod, d , I₀ 1m 2 2 3     (c) Uniform ring, d = R, 2 2 2 2 0 cm I I md mR mR 2mR (d) Uniform disc, d = R, 2 2 2 2 0 cm mR 3 I I md mR mR 2 2     

FEW MISCELLENOUS QUESTIONS OF S.H.M.

Example-1 :

A spring is connected with a solid sphere as shown in the figure. k

M, R

Surface is rough enough to prevent slipping. Find out time period of small oscillations ?

Solution :

We should solve it by energy method

Consider any situation whenits speed is ‘v’ angular speed is ‘’ and extension in spring is ‘x’. E = K + U = constant 1mv2 1I 2 1kx2 2 2 2     k k v x I 2 2 1 1 m v kx 2 2

  = constant (where m_I = m[1+I/mR²])

Now, dE 0 dt  I 1 dv 1 dx m 2v k2x 0 2 dt 2 dt    2 I 2 d x m v kxv 0 dt    2 2 I d x k x 0 dt m    2 I k m    _{(compare with} 2 2 2 d x x 0 dt    ) I m 2 T 2 R       2 I 2 2 mR 7 5 m m 1 m mR 5      _  _     7m T 2 5k   

Example-2 :

If the above arrangement is changed slightly

k

M, R k

by adding one more spring as per the diagram. Please find time period of small oscillation ?

Solutoin :

 E = K + U 1mv2 1I 2 1kx2 1k 2x

 

2 2 2 2 2      I 2 2 2 1 1 4 m v kx kx 2 2 2    x 2x v I 2





2 1 1 m v 5k x 2 2  

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From the previous results





I eq m 7m T 2 2 k 5 5k     _2v v As per pure rolling concept

7m T 2

25k  

Example-3 :

Please refer to given arrangement as shown in figure. Find out time period of small displacement?

k k O mass = m, length =

Solution :

kx _k O kx cos kx cos kx x =  2

For very small ‘’ sin  = tan  =  cos  = 1

Restoring torque about ‘O’

r kx cos kx cos 2 2     _    _    

_r = Force × perpendicular distance

r I k k 2 2 2 2           _{ } _  _ _{ }           x 2  2 2 2 2 m d k 12 dt 2      _ _     2 2 d 6k 0 dt m     _{ }    compare it with 2 2 2 d 0 dt      6k m   m T 2 6k  

Note the above question can also be asked by the following form.

R

R O

Example-4

Two masses are attached with an ideal spring as shown in figure. Find out time period of blocks for small oscillations? m2 k m1 (Smooth surface)

Solution :

Please refer to FBD as shown

m2 m1 m2 m1 kx kx x = x + x_{1 2} FBD of m₁ m1 r F  kx 2 1 1 2 d x m kx dt   2 1 2 1 d x k x dt  m ...(1) FBD of m₂ : 2 r F  kx 2 2 2 2 d x m kx dt   2 2 2 2 d x k x dt  m ...(2) From (1) & (2)  (1) + (2);





2 1 2 2 1 2 d x 1 1 x x k x dt m m      _  _   2 2 r d x k x dt m    2 2 r d x k x 0 dt m    where r 1 2 1 2 m m m m m 

 also called reduced mass comparing with standard equation of SHM

2 r r m k T 2 m k      _,





1 2 1 2 m m T 2 k m m   

The above question can be represented as

m2 k

m1

mr k

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Example-5

A block is connected with an elastic string as shown in figure.

m Smooth

Find out time period of small oscillation ? Given Young’s modulus of string is Y, A = Cross sectional area of wire and

 = initial length of the wire.

Solution :

The string/wire will only perform SHM in half of its

x=A m m x=0 m t= T 4 t= T 2

motion i.e. when it is in extension. But its motion cannot be complete SHM as when it moves towards its length after mean position then its motion will not be SHM. Elastic constant k YA



& half time period T 1 2 m 2 2 k         T m 2   YA 

Example-6

If in the above question a spring of spring constant ‘k’ is

m

k _{wire, Y,A,}

smooth surface

attached to other end of the mass as shown in the figure, then find its time period?

Solution :

This case is of the combination of two SHM.

Part-1 : When both spring and wire are in parallel

eq m T 2 k   x=0 m k m m k t=T 4 t= T 2 t=0 x=–A k1 k =1 YA 

both spring and string are engaged

where k eq = k + k1 = kYA  Part-2 : m T ' 2 k   m m t= T 4 t= T 2 x=+A only spring is engaged

Total Time period T_net T T ' 2 2    1 m m 2 2 YA 2 k k             _     _{ }    