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 = – ω²x ……………….. (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

Introduction to SHM Video [In HINDI+ENGLISH Mix Language]

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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 P_o 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 ……….. (i)

Differentiating equation (i), we get

Or ……… (ii)

Differentiating equation (ii), we get

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.

Relation Between SHM and Circular Motion Video

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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.
v_max = ωA
At extreme position (x = A), velocity is minimum.
v_min = 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.
a_min = zero
At extreme position (x = A), acceleration is maximum.
a_max = ω²A

Graph of acceleration (a) vs displacement (x)

a = – ω²x

Slope of the graph = – ω²

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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 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 =