Simple Harmonic Motion
Periodic Motion:
(i) When a body moves on a definite path and repeats its motion again and again after a definite interval of time, then its motion is called periodic motion.
The constant time interval after which it repeats its motion called its time period of motion, T.
Example : Motion of earth around Sun ( T=365.25 days), Motion of minutes hand in clock (T= 1hour), Motio of a simple pendulum in free space, etc.
So, for motion of a body to be periodic,
* Body must move on a definite path.
* It must repeat its motion after a definite time interval.
Oscilatory Motion:
When a body makes to and fro motion about a fixed point, on a almost straight line path and repeats its motion after a definite time interval, then the motion of the body is a vibratory motion or oscillatory motion
* It must have periodic motion
* The body must move to & fro about a fixed point on a almost straight line path.
The point about which the body makes to and fro motion is called its mean position.
Example:
(i) Motion of a bob of a simple pendulum if displacement is small, (ii) Motion of a mass attached to a spring,
(iii) Motion of a block, moving with constant speed, on a smooth horizontal floor between two rigid walls.
* Every oscillatory motion must be a periodic motion but every periodic motion is not oscillatory.
Simple Harmonic Motion
When a body makes oscillatory motion about a fixed mean position such that the acceleration of the body at any instant is directly proportional to its displacement from the mean position & is always directed towards the mean position
a ∝ –x
or a = – ω2 x ……………….. (1)
Where, ω (const) = angular frequency
According to Newtons second law –
F = ma
or F = –mω2x
F = –k x ……………….. (2)
i.e. F ∝ –x
Where, k = Force constant
Since, the force is always directed towards the mean position it is called restoring force.
- Note :The necessary and sufficient condition for a motion of a particle to be S.H.M. that, force acting on it must be directed towards a fixed point called mean (equilibrium) position and magnitude of the force must be proportional to the displacement of the particle from the equilibrium position.
- Note: For SHM F = –k x
SHM and uniform circular motion
Consider a particle moving on a circle of radius A with a constant angular speed ω as shown in figure.
Suppose the particle is at point Po on the circle, at t = 0. The radius OP make an angle θ = (ωt + φ) with the X-axis at time t. Drop a perpendicular PM on X-axis and drop a perpendicular PN on Y-axis. Let us trace the path followed by the points M and N with time, as particle P moves with constant angular speed on the circular path. We find, both the points M and N makes oscillatory motion, about the mean position at the center of the circle.
For point ‘N’:
Displacement of the point N from mean position is ON = Y ,
In ΔOPN 
Or 
Or 
……….. (i)
Differentiating equation (i), we get
Or 
……… (ii)
Differentiating equation (ii), we get 
Or 
Or 
……………(iii)
Thus, a ∝ – Y
⇒ Thus the point N (projection of particle P, on Y- axis executes SHM.
Similarly, for point ‘M’, which is the projection of point P on X-axis, we get
Displacement at t = t, is
X = A cos (ωt + f0) …………(iv)
…………(v)
and 
…………………..(vi)
Thus, point ‘M’ also makes SHM.
Characteristics of SHM:
Note : In the figure shown, path of the particle is on a straight line.
(a) Displacement – It is defined as the distance of the particle from the mean position at that instant.
Displacement in SHM at time t is given by x = A sin (ωt + φ)
(b) Amplitude – It is the maximum value of displacement of the particle from its equilibrium position.
Amplitude = 
[distance between extreme points/position]
It depends on energy of the system.
(c) Angular Frequency (ω) : ω= 2π/T = 2πf and its units is rad/sec.
(d) Frequency (f) : Number of oscillations completed in unit time interval is called frequency of oscillations, f = 1/T = ω/2π , its units is sec–1 or Hz.
(e) Time period (T) : Smallest time interval after which the oscillatory motion gets repeated is called time period, 
Example. For a particle performing SHM, equation of motion is given as 
. Find the time period.
Sol. Given
Or 
or ω2 = 4
ω = 2
Time period; 
(f) Phase : The physical quantity which represents the state of motion of particle (eg. its position and direction of motion at any instant).
The argument (ωt + φ) of sinusoidal function is called instantaneous phase of the motion.
(g) Phase constant ( φ ) : Constant φ in equation of SHM is called phase constant or initial phase.
It depends on initial position and direction of velocity.
(h) Velocity(v) : It is the rate of change of particle’s displacement w.r.t time at that instant.
Let the displacement from mean position is given by
x = A cos (ωt + φ)
Velocity, v = – Aw cos (ωt + φ)
or, 
At mean position (x = 0), velocity is maximum.
vmax = ωA
At extreme position (x = A), velocity is minimum.
vmin = zero
Graph of speed (v) vs displacement (x):
Graph would be an ellipse.
(i) Acceleration : It is the rate of change of particle’s velocity w.r.t. time at that instant.
Acceleration, a = – ω2A cos (ωt + φ)
a = – ω2x
NOTE: Negative sign shows that acceleration is always directed towards the mean postion.
At mean position (x = 0), acceleration is minimum.
amin = zero
At extreme position (x = A), acceleration is maximum.
amax = ω2A
Graph of acceleration (a) vs displacement (x)
Graph of acceleration (a) vs displacement (x)
a = – ω2x
Slope of the graph = – ω2
Example: The equation of particle executing simple harmonic motion is 
. Write down the amplitude, time period and maximum speed. Also find the velocity at t = 1 s.
Solution: Comparing with equation x = A sin (wt + δ), we see that
the amplitude = 5 m,
and time period = = 
The maximum speed = A w = 5 m × π s–1 = 5π m/s.
The velocity at time t =
= A w cos (wt + δ)
At t = 1 s,
v = (5 m) (π s–1) cos = 
Example: A particle executing simple harmonic motion has angular frequency 6.28 s–1 and ampitude 10 cm. Find (a) the time period, (b) the maximum speed, (c) the maximum acceleration, (d) the speed when the displacement is 6 cm from the mean position, (e) the speed at t = 1/6 s assuming that the motion starts from rest at t = 0.[HCV]
Solution: (a) Time period = 
= 1 s.
(b) Maximum speed = Aw = (0.1 m) (6.28 s–1)
= 0.628 m/s.
(c) Maximum acceleration = Aw2
= (0.1 m) (6.28 s–1)2
= 4 m/s2.
(d) 
w = (6.28 s–1)
= 50.2 cm/s.
(e) At t = 0, the velocity is zero i.e., the particle is at an extreme. The equation for displacement may be written as
x = A coswt.
The velocity is v = – A w sin wt.
At t = 1/6 s, v = – (0.1 m) (6.28 s–1) sin
= ( – 0.628 m/s) sin = 54.4 cm/s.
Example: A particle starts from mean position and moves towards positive extreme as shown.
Find the equation of the SHM. Amplitude of SHM is A.
Solution: General equation of SHM can be written as x = A sin (wt + φ)
At t = 0, x = 0
So, 0 = A sinφ
⇒ φ= 0, π
Also; at t = 0, v = +ve
∴ Aω cosφ = +ve
or, φ = 0
Hence, if the particle is at mean position at t = 0 and is moving towards +ve extreme, then the equation of SHM is given by x = A sin ωt
Similarly
for φ = π
From equation of SHM is x = A sin(ωt + π)
or, x = -A sinωt
NOTE:
If mean position is not at the origin, then we can replace x by x – x0 and the eqn. becomes x – x0 = -A sinωt, where x0 is the position co-ordinate of the mean position.
Example: A particle is performing SHM of amplitude “A” and time period “T”. Find the time taken by the particle to go from 0 to A/2.
Solution: Let equation of SHM be x = A sin wt
when x = 0 , t = 0
when x = A/2 ; putting the value of x ,in its eq. we get
A/2 = A sin wt
or sin wt = 1/2 ⇒ wt = π/6
t = π/6ω ⇒ t = T/12
Hence , time taken is T/12, where T is time period of SHM.
Example: A particle of mass 2 kg is moving on a straight line under the action force F = (8 – 2x) N. It is released at rest from x = 6 m.
(a) Is the particle moving simple harmonically.
(b) Find the equilibrium position of the particle.
(c) Write the equation of motion of the particle.
(d) Find the time period of SHM.
Solution: F = 8 – 2x or F = –2(x – 4)
for equilibrium position F = 0 ⇒ x = 4 is equilibrium position
Hence the motion of particle is SHM with force constant 2 and equilibrium position x = 4.
(a) Yes, motion is SHM.
(b) Equilibrium position is x = 4
(c) At x = 6 m, particle is at rest i.e. it is one of the extreme position
Hence amplitude is A = 2 m and initially particle is at the extreme position.
∴ Equation of SHM can be written as
x – 4 = 2 cos ωt , where ω = 
i.e. x = 4 + 2 cos t
(d) Time period, T = 2π/ω = 2π sec.