Simple Harmonic Motion And Oscillation

PhysicsOscillationsFor NEET aspirants

Simple harmonic motion (SHM) is the to-and-fro motion of a particle about a mean position under a restoring force that is proportional to the displacement and always points back to the mean position: $F = - k x$ . Every such motion follows $x = A sin (ω t + ϕ)$ with $ω = k / m$ and $T = 2 π / ω$ . This page builds simple harmonic motion step by step: oscillatory motion, the equation of SHM, velocity and acceleration, graphs, the reference circle (phasor) method and energy. It is a scoring chapter in both JEE Main and NEET.

On this page1Periodic & oscillatory2Restoring force3Equation of SHM4Velocity & acceleration5Graphs6Reference circle7Energy

Key Formulas - Quick Reference

★ Must learnCondition for SHM: $F = - k x$ , or $a = - ω^{2} x$
★ Must learnDisplacement: $x = A sin (ω t + ϕ)$ ; phase $= ω t + ϕ$
$ω = k / m = 2 π f = \frac{2 π}{T}$ , so $T = 2 π m / k$
★ Must learnVelocity: $v = A ω cos (ω t + ϕ) = \pm ω A^{2} - x^{2}$ ; $v_{m a x} = A ω$ at the mean position
Acceleration: $a = - ω^{2} x$ ; $∣ a ∣_{m a x} = ω^{2} A$ at the extremes
$v$ against $x$ is an ellipse: $(\frac{x}{A})^{2} + (\frac{v}{A ω})^{2} = 1$
Phase order: $v$ leads $x$ by $\frac{π}{2}$ ; $a$ leads $v$ by $\frac{π}{2}$ ( $a$ is opposite to $x$ )
Energy: $U = \frac{1}{2} k x^{2}$ , $K = \frac{1}{2} k (A^{2} - x^{2})$ , $E = \frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2}$
Time averages: $⟨ K ⟩ = ⟨ U ⟩ = \frac{E}{2}$ ; $K$ and $U$ oscillate at $2 f$
Reference circle: $t = \frac{Δ θ}{ω}$ ; $O \to A /2$ takes $T /12$ , $A /2 \to A$ takes $T /6$

1. Periodic and Oscillatory Motion

1.1 Periodic motion

A motion that repeats itself along the same path after equal intervals of time is called periodic motion. The smallest such interval is the time period $T$ . The path can be a straight line, a circle, an ellipse or any other curve. Example: the Earth going round the Sun repeats every year.

1.2 Oscillatory motion

A to-and-fro motion about a stable equilibrium position is called oscillatory motion. The force (or torque) that always acts towards the equilibrium position is the restoring force (restoring torque). Oscillatory motion need not be periodic and need not have fixed extreme positions. If energy is conserved, the oscillation is also periodic, like the pendulum of a wall clock. If a resistive force such as air drag removes energy, the total mechanical energy falls and the motion is a damped oscillation, which finally stops.

Periodic motion

Repeats after a fixed time $T$ . Need not be to-and-fro. Example: Earth around the Sun, a fan blade.

Oscillatory motion

To-and-fro about a stable equilibrium, driven by a restoring force. Need not repeat exactly. Example: a damped swing.

Figure 1: Every SHM is periodic and oscillatory, but not every periodic or oscillatory motion is SHM. SHM is the special case with restoring force

F = - k x

2. The Oscillatory Equation $F = - k x^{n}$

Consider a particle free to move along the $x$ -axis under a force $F = - k x^{n}$ , where $k$ is a positive constant and $x$ is the displacement from $x = 0$ . Whether it oscillates depends only on $n$ .

Value of $n$	Direction of force	Result
Even ( $0, 2, 4, \dots$ )	Always along $- x$ , whether $x$ is positive or negative	Not oscillatory: released anywhere except $x = 0$ , the particle keeps moving along $- x$
Odd ( $1, 3, 5, \dots$ )	Along $- x$ for $x > 0$ , along $+ x$ for $x < 0$ , zero at $x = 0$	Oscillates about the stable equilibrium $x = 0$ ; the force is a restoring force
$n = 1$	$F = - k x$ : restoring and proportional to $x$	Simple harmonic motion

Figure 2:

F = - k x^{n}

. Red arrows show the force on each side of

x = 0

. Only

n = 1

gives SHM; odd

n > 1

oscillates but is not SHM; even

n

does not oscillate.

★ Must learnSimple harmonic motion: if the restoring force (or torque) on a body is directly proportional to its displacement (or angular displacement) from the mean position and is always directed towards the mean position, the motion is simple harmonic. It is the simplest form of oscillatory motion:

F = - k x

Here

k

is the SHM constant (force constant), in

N m^{- 1}

Exam Trick

The one-line SHM test. Rewrite the force as $F = - k (x - x_{0})$ . If you can, the motion is SHM about the mean position $x = x_{0}$ with $ω = k / m$ . A constant term only shifts the mean position. If the coefficient of $x$ is positive ( $F = + k x + c$ ), the force pushes the particle away: no SHM. Solved Examples 1 and 6 use this test.

Figure 3: Two questions decide it. A constant term in

F

only moves the mean position to

x_{0}

; the sign of the

x

coefficient decides between SHM (

F = - 3 x + 3

) and run-away motion (

F = 3 x - 3

Key idea

SHM needs two things together: force towards the mean position and force proportional to the distance from it.

3. Types of SHM

Linear SHM

The particle moves to and fro along a straight line about a mean point $M$ . If $A$ and $B$ are the extreme positions, $A M = M B =$ amplitude. Example: a block on a spring.

Angular SHM

The body rotates to and fro about a fixed axis; the restoring torque is proportional to the angular displacement, $τ = - k θ$ . Example: a pendulum, a torsional pendulum.

This page treats linear SHM; the same ideas carry over to angular SHM with $x \to θ$ , $F \to τ$ and $m \to I$ (covered in the pendulums concept).

4. Motion of the Particle in One Cycle

Displace the particle from the mean position and release it. The force $- k x$ pulls it back. By the time it reaches the mean position it has gained kinetic energy, so it overshoots, stops somewhere on the other side and is pulled back again. Take the particle at the mean position $O$ at $t = 0$ moving towards $+ A$ with speed $v_{0}$ . One period splits into four quarter-cycles:

$O \to + A$ : the force towards $O$ grows, so the acceleration towards $O$ grows while the speed falls; the particle stops momentarily at $+ A$ .
$+ A \to O$ : it starts from rest; the force towards $O$ shrinks as it approaches $O$ , so the acceleration falls while the speed grows; it reaches $O$ with the same speed $v_{0}$ .
$O \to - A$ : same as the first part, on the other side.
$- A \to O$ : same as the second part, on the other side.

Motion	Velocity: direction, magnitude	Acceleration: direction, magnitude
$O \to + A$	towards $+ A$ , decreasing	towards $O$ , increasing
$+ A \to O$	towards $O$ , increasing	towards $O$ , decreasing
$O \to - A$	towards $- A$ , decreasing	towards $O$ , increasing
$- A \to O$	towards $O$ , increasing	towards $O$ , decreasing

Figure 4: Moving from

- A

+ A

. Velocity (orange) is largest at the mean position; acceleration (red) always points to

O

and is largest at the extremes.

Key idea

Acceleration always points to the mean position. Velocity points to the mean position only on the way back.

5. Characteristics of SHM

Term	Meaning	Symbol and unit
Mean position	Position where the net force on the particle is zero (stable equilibrium)	$x = 0$ (or $x_{0}$ )
Extreme position	Position where the speed of the particle is zero	$x = \pm A$
Displacement	Position of the particle measured from the mean position at that instant	$x$ , m
Amplitude	Maximum displacement from the mean position: extreme position minus mean position. Depends on the energy given to the system	$A$ , m
Time period	Smallest time after which the motion repeats	$T = \frac{2 π}{ω}$ , s
Frequency	Number of complete oscillations per second	$f = \frac{1}{T} = \frac{ω}{2 π}$ , Hz ( $s^{- 1}$ )
Angular frequency	Rate of change of phase	$ω = 2 π f$ , $rad s^{- 1}$

Amplitude depends on energy; period does not. For a given spring and mass, pulling harder gives a bigger $A$ but the same $T = 2 π m / k$ . This is why SHM clocks keep time.

6. Equation of SHM

6.1 The differential equation

The necessary and sufficient condition for SHM is $F = - k x$ .
Newton's second law: $ma = - k x$ , that is $m \frac{d ^{2} x}{d t ^{2}} = - k x$ .
Divide by $m$ and write $ω^{2} = \frac{k}{m}$ :
$\frac{d ^{2} x}{d t ^{2}} + ω^{2} x = 0$
This is the differential equation of SHM. Any quantity that obeys it oscillates simple harmonically.
Its general solution is
$x = A sin (ω t + ϕ)$
where $A$ is the amplitude, $ω = k / m$ is the angular frequency and $k$ is the SHM constant.

★ Must learnPhase and phase constant. The angle

(ω t + ϕ)

is the phase: it fixes both the position and the direction of motion at time

t

. The phase constant (initial phase)

ϕ

is the phase at

t = 0

; it depends on the starting position and the direction of the starting velocity.

6.2 Finding $ϕ$ from the starting point

Put $t = 0$ : $x_{0} = A sin ϕ$ gives two possible angles; the sign of $v_{0} = A ω cos ϕ$ picks one of them. Particle at $O$ moving towards $+ A$ : $ϕ = 0$ , so $x = A sin ω t$ . Particle at $+ A$ at $t = 0$ (moving towards $O$ ): $ϕ = π /2$ , so $x = A sin (ω t + π /2) = A cos ω t$ . Here $π /2$ is the only possible phase.

Start ( $t = 0$ )	Direction	$ϕ$	Equation
$x = 0$	towards $+ A$	$0$	$x = A sin ω t$
$x = + A$	(at rest)	$\frac{π}{2}$	$x = A cos ω t$
$x = 0$	towards $- A$	$π$	$x = - A sin ω t$
$x = - A$	(at rest)	$\frac{3 π}{2}$	$x = - A cos ω t$
$x = + A /2$	towards $+ A$	$\frac{π}{6}$	$x = A sin (ω t + π /6)$
$x = + A /2$	towards $O$	$\frac{5 π}{6}$	$x = A sin (ω t + 5 π /6)$

Figure 5: Phase dial for

x = A sin (ω t + ϕ)

. Measure

ϕ

clockwise from the top; the dot's shadow on the horizontal line is the starting position.

Mean position not at the origin. If the mean position is at $x = x_{0}$ , replace $x$ by $x - x_{0}$ :

x - x_{0} = A sin (ω t + ϕ)

The cosine form $x = A cos (ω t + ϕ^{'})$ used by NCERT is the same motion with $ϕ^{'} = ϕ - π /2$ ; use whichever makes $ϕ$ simplest.

7. Velocity and Acceleration in SHM

7.1 Velocity

Velocity is the rate of change of displacement. From $x = A sin (ω t + ϕ)$ :

v = \frac{d x}{d t} = A ω cos (ω t + ϕ)

Using $cos = 1 - sin^{2}$ and $sin (ω t + ϕ) = x / A$ :

v = \pm ω A^{2} - x^{2}

At the mean position ( $x = 0$ ) the speed is maximum, $v_{m a x} = A ω$ . At the extremes ( $x = \pm A$ ) it is zero. Squaring and rearranging gives

\frac{x ^{2}}{A ^{2}} + \frac{v ^{2}}{A ^{2} ω ^{2}} = 1

so the $v$ - $x$ graph is an ellipse (a circle if the axes are scaled so that $A ω$ and $A$ are drawn equal). For $v \geq 0$ alone it is a half ellipse.

7.2 Acceleration

Acceleration is the rate of change of velocity:

a = \frac{d v}{d t} = - A ω^{2} sin (ω t + ϕ) = - ω^{2} x

The negative sign shows that acceleration is always directed towards the mean position. It is zero at the mean position and maximum, $∣ a ∣_{m a x} = ω^{2} A$ , at the extremes. The $a$ - $x$ graph is a straight line through the origin with slope $- ω^{2}$ .

Figure 6: Left,

v

against

x

is an ellipse,

(x / A)^{2} + (v / A ω)^{2} = 1

. Right,

a = - ω^{2} x

is a straight line of slope

- ω^{2}

Exam Trick

Two positions, two speeds: find $ω$ and $A$ at once. From $v^{2} = ω^{2} (A^{2} - x^{2})$ at two points,

ω^{2} = \frac{v _{1}^{2} - v _{2}^{2}}{x _{2}^{2} - x _{1}^{2}}, A^{2} = x_{1}^{2} + \frac{v _{1}^{2}}{ω ^{2}}

Also, the ratio $\frac{a _{m a x}}{v _{m a x}} = ω$ and $\frac{v _{m a x}^{2}}{a _{m a x}} = A$ . These relations hold for any SHM, whatever the starting phase.

8. Graphs of Displacement, Velocity and Acceleration

Take $x = A sin ω t$ . Then $v = A ω cos ω t = A ω sin (ω t + π /2)$ and $a = - A ω^{2} sin ω t = A ω^{2} sin (ω t + π)$ .

Time $t$	$0$	$T /4$	$T /2$	$3 T /4$	$T$
Displacement $x$	$0$	$A$	$0$	$- A$	$0$
Velocity $v$	$A ω$	$0$	$- A ω$	$0$	$A ω$
Acceleration $a$	$0$	$- ω^{2} A$	$0$	$ω^{2} A$	$0$

Figure 7: For

x = A sin ω t

v

peaks a quarter period before

x

, and

a

is always opposite to

x

. Dashed guides mark

T /4, T /2, 3 T /4, T

Displacement, velocity and acceleration all vary harmonically with time, with the same period $T$ .
The maximum velocity is $ω$ times the amplitude: $v_{m a x} = A ω$ .
The maximum acceleration is $ω^{2}$ times the amplitude: $a_{m a x} = ω^{2} A$ .
Velocity is ahead of displacement by a phase angle of $π /2$ .
Acceleration is ahead of velocity by $π /2$ , so it is $π$ ahead of (opposite to) displacement.
The relations $v = \pm ω A^{2} - x^{2}$ and $a = - ω^{2} x$ hold for every form of the equation of $x$ .

Quick Recall: tap to check

At which position is the speed maximum and the acceleration zero?

At the mean position (

x = 0

What is the phase difference between displacement and acceleration?

π

: acceleration is always opposite to displacement.

What is the shape of the

v

x

graph?

An ellipse with semi-axes

A

and

A ω

x = A cos ω t

, what is

v

t = T /4

v = - A ω sin (π /2) = - A ω

(maximum speed, moving towards

- x

9. SHM as the Projection of Uniform Circular Motion

Let a point $Q$ move clockwise on a circle of radius $A$ with constant angular velocity $ω$ . Drop a perpendicular from $Q$ onto the horizontal diameter; its foot $P$ moves to and fro between $- A$ and $+ A$ in simple harmonic motion. If the radius $O Q$ makes an angle $ϕ$ with the upward vertical at $t = 0$ , then at time $t$ the angle is $ω t + ϕ$ and

x (t) = A sin (ω t + ϕ)

Because the foot moves on the horizontal diameter, this picture is called the horizontal phasor. The foot of the perpendicular on the vertical diameter also does SHM, $y (t) = A cos (ω t + ϕ)$ (the vertical phasor); the two differ in phase by $π /2$ .

Figure 8:

Q

moves clockwise on a circle of radius

A

at constant

ω

; its shadow

P

does SHM,

x = A sin (ω t + ϕ)

. The horizontal part of

Q

's velocity is

P

's velocity.

9.1 Problem-solving strategy (horizontal phasor)

Draw a circle of radius equal to the amplitude of the SHM.
Imagine a particle going round it clockwise with the same $ω$ as the SHM.
Mark the starting point: its shadow is the starting position, and the side of the circle is chosen so that the horizontal component of its velocity matches the starting direction. The angle from the upper vertical is $ϕ$ .
The horizontal component of the circling particle's velocity, $A ω cos (ω t + ϕ)$ , is the SHM velocity.
The horizontal component of its centripetal acceleration $ω^{2} A$ is the SHM acceleration, $a (t) = - ω^{2} A sin (ω t + ϕ)$ .
Time between any two states: $t = \frac{Δ θ}{ω} = \frac{Δ θ}{2 π} T$ , where $Δ θ$ is the angle turned on the circle.

Figure 9: Times measured from the mean position, found from

t = Δ θ / ω

. Moving the last half of the distance takes twice as long as the first half.

Exam Trick

Memorise the time ladder. From the mean position: to $A /2$ takes $T /12$ ( $3 0^{\circ}$ ), to $A / 2$ takes $T /8$ ( $4 5^{\circ}$ ), to $3 A /2$ takes $T /6$ ( $6 0^{\circ}$ ), to $A$ takes $T /4$ . So $A /2 \to A$ takes $T /4 - T /12 = T /6$ . The particle is slowest near the extremes, so equal distances there take longer.

JEE Advanced

Two SHMs of the same $ω$ : their phase difference stays constant, so their radii on the reference circle turn together like the hands of a rigid clock. Two particles meet when their shadows coincide; the separation $x_{1} - x_{2}$ is itself an SHM of amplitude $A_{1}^{2} + A_{2}^{2} - 2 A_{1} A_{2} cos δ$ , which gives the maximum separation directly (Solved Example 10).

10. Energy in SHM

10.1 Kinetic energy

K = \frac{1}{2} m v^{2} = \frac{1}{2} m A^{2} ω^{2} cos^{2} (ω t + ϕ) = \frac{1}{2} m ω^{2} (A^{2} - x^{2})

Since $m ω^{2} = k$ : $K = \frac{1}{2} k (A^{2} - x^{2})$ . $K_{m a x} = \frac{1}{2} k A^{2}$ at $x = 0$ and $K_{m i n} = 0$ at $x = \pm A$ . Because $cos^{2}$ repeats every half period, the frequency of kinetic energy is twice the frequency of SHM.

10.2 Potential energy

Work done by $F = - k x$ in a small displacement $d x$ : $d W = F d x = - k x d x$ .
From $x = 0$ to $x$ : $W = \int_{0}^{x} (- k x) d x = - \frac{1}{2} k x^{2}$ .
For a conservative force the change in potential energy is the negative of the work: $U (x) - U_{M.P.} = \frac{1}{2} k x^{2}$ .
Choose $U = 0$ at the mean position:
$U (x) = \frac{1}{2} k x^{2} = \frac{1}{2} m ω^{2} x^{2} = \frac{1}{2} m ω^{2} A^{2} sin^{2} (ω t + ϕ)$

10.3 Total mechanical energy

★ Must learn

E = K + U = \frac{1}{2} m ω^{2} A^{2} = \frac{1}{2} k A^{2} = constant

The total energy is proportional to the square of the amplitude and does not depend on time or position.

Figure 10:

U = \frac{1}{2} k x^{2}

and

K = \frac{1}{2} k (A^{2} - x^{2})

always add to the constant

E = \frac{1}{2} k A^{2}

. They are equal at

x = \pm A / 2

Figure 11: For

x = A sin ω t

U = E sin^{2} ω t

K = E cos^{2} ω t

. Energies repeat every

T /2

, so they oscillate at twice the SHM frequency.

Quantity	Averaged over time (one period)	Averaged over displacement ( $0$ to $A$ )
Kinetic energy	$\frac{1}{4} k A^{2} = \frac{E}{2}$	$\frac{1}{3} k A^{2} = \frac{2 E}{3}$
Potential energy	$\frac{1}{4} k A^{2} = \frac{E}{2}$	$\frac{1}{6} k A^{2} = \frac{E}{3}$

Note: $U_{m i n} = U_{M.P.}$ , which is not always zero. The zero of potential energy is a choice; only $U - U_{M.P.} = \frac{1}{2} k x^{2}$ is fixed. For a vertical spring, gravity and the spring together still give $\frac{1}{2} k x^{2}$ measured from the new mean position.

JEE Advanced

SHM from a potential-energy curve. Near a minimum $x_{0}$ of any $U (x)$ , Taylor expansion gives $U \approx U (x_{0}) + \frac{1}{2} U^{''} (x_{0}) (x - x_{0})^{2}$ , so the force is $F = - U^{''} (x_{0}) (x - x_{0})$ . Small oscillations about any stable equilibrium are therefore SHM with

ω = \frac{U ^{''} ( x _{0} )}{m}

This is the fastest route when a question gives $U (x)$ instead of $F (x)$ (Solved Example 16).

Figure 12:

U = U_{0} (1 - cos a x)

(solid) and its parabola

\frac{1}{2} U_{0} a^{2} x^{2}

(dashed). Near the minimum they agree to 1% at

∣ a x ∣ = 0.35

and 2% at

0.5

, so small oscillations are SHM with

ω = a U_{0} / m

(Solved Example 16). About the maxima the force pushes away: no oscillation.

Quick Recall: tap to check

Where is kinetic energy equal to potential energy?

x = \pm A / 2

, where each equals

E /2

If the amplitude doubles, what happens to the total energy?

It becomes four times, since

E \propto A^{2}

What is the period of the kinetic energy of an SHM of period

T

T /2

Figure 13: The whole concept on one screen. Revise it branch by branch; every numerical on this page uses one of these six ideas.

11. Solved Examples

Solved Example 1

Describe the motion of a particle acted upon by a force (i)

F = 3 x + 3

(ii)

F = - 3 x - 3

(iii)

F = - 3 x + 3

(iv)

F = 3 x - 3

Solution:

For each force, find the mean position ( $F = 0$ ) and check whether the force on either side points towards it.

(i) $F = 3 x + 3 = 0 \Rightarrow x = - 1$ . At $x = 0$ : $F = + 3 N$ , away from $x = - 1$ . At $x = - 2$ : $F = - 3 N$ , again away. Not SHM (the particle runs away from $x = - 1$ ).

(ii) $F = - 3 x - 3 = - 3 (x + 1)$ , mean position $x = - 1$ . At $x = 0$ : $F = - 3 N$ (towards); at $x = - 2$ : $F = + 3 N$ (towards). SHM about $x = - 1$ .

(iii) $F = - 3 x + 3 = - 3 (x - 1)$ , mean position $x = 1$ . At $x = 0$ : $F = + 3 N$ (towards); at $x = 2$ : $F = - 3 N$ (towards). SHM about $x = 1$ .

(iv) $F = 3 x - 3 = 3 (x - 1)$ , mean position $x = 1$ . At $x = 0$ : $F = - 3 N$ (away); at $x = 2$ : $F = + 3 N$ (away). Not SHM.

Answer: (ii) and (iii) are SHM; (i) and (iv) are not. Shortcut: SHM only when the coefficient of $x$ is negative.

Solved Example 2

A particle starts from the mean position and moves towards the positive extreme. Find the equation of the SHM. The amplitude is

A

Solution:

General equation: $x = A sin (ω t + ϕ)$ . At $t = 0$ , $x = 0$ : $0 = A sin ϕ$ , so $ϕ = 0$ or $π$ (with $ϕ \in [0, 2 π)$ ). Also $v = A ω cos ϕ$ must be positive, so $ϕ = 0$ .

Answer: $x = A sin ω t$ . Similarly, if it starts from the mean position towards the negative extreme, $ϕ = π$ and $x = A sin (ω t + π) = - A sin ω t$ .

Solved Example 3

A particle in SHM is at

x = A /2

t = 0

and is moving towards the mean position. Write the equation of the SHM.

Solution:

$x = A sin (ω t + ϕ)$ . At $t = 0$ : $\frac{A}{2} = A sin ϕ$ , so $ϕ = 3 0^{\circ}$ or $15 0^{\circ}$ . The velocity $A ω cos ϕ$ is negative (moving towards $O$ from the positive side), so $cos ϕ < 0$ and $ϕ = 15 0^{\circ} = 5 π /6$ .

Answer: $x = A sin (ω t + \frac{5 π}{6})$ .

Solved Example 4

The equation of a particle executing SHM is

x = (5 m) sin [(π s^{- 1}) t + \frac{π}{3}]

. Write down the amplitude, time period and maximum speed. Also find the velocity at

t = 1 s

Solution:

Comparing with $x = A sin (ω t + ϕ)$ : amplitude $A = 5 m$ , $ω = π s^{- 1}$ .

Time period $T = \frac{2 π}{ω} = \frac{2 π}{π} = 2 s$ . Maximum speed $= A ω = 5 \times π = 5 π m s^{- 1}$ .

Velocity: $v = A ω cos (ω t + ϕ)$ . At $t = 1 s$ :

v = (5) (π) cos (π + \frac{π}{3}) = 5 π (- \frac{1}{2}) = - \frac{5 π}{2} m s^{- 1}

Answer: $A = 5 m$ , $T = 2 s$ , $v_{m a x} = 5 π m s^{- 1}$ , $v (1 s) = - 2.5 π \approx - 7.85 m s^{- 1}$ .

Solved Example 5

A particle executing SHM has angular frequency

6.28 s^{- 1}

and amplitude

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

Solution:

(a) $T = \frac{2 π}{ω} = \frac{6.28}{6.28} = 1 s$ .

(b) $v_{m a x} = A ω = (0.1 m) (6.28 s^{- 1}) = 0.628 m s^{- 1} = 62.8 cm s^{- 1}$ .

(c) $a_{m a x} = A ω^{2} = (0.1) (6.28)^{2} \approx 3.9 m s^{- 2} \approx 4 m s^{- 2}$ .

(d) $v = ω A^{2} - x^{2} = (6.28) 1 0^{2} - 6^{2} cm s^{- 1} = 6.28 \times 8 = 50.2 cm s^{- 1}$ .

(e) Starting from rest means starting at an extreme, so $x = A cos ω t$ and $v = - A ω sin ω t$ . At $t = \frac{1}{6} s$ : $v = - (0.628) sin (\frac{6.28}{6}) = - 0.628 sin \frac{π}{3} = - 0.544 m s^{- 1}$ .

Answer: (e) speed $= 54.4 cm s^{- 1}$ , directed towards the mean position.

Solved Example 6

A particle of mass

2 kg

moves on a straight line under the force

F = (8 - 2 x) N

. It is released at rest from

x = 6 m

. (a) Is the motion simple harmonic? (b) Find the equilibrium position. (c) Write the equation of motion. (d) Find the time period.

Solution:

Rewrite: $F = - 2 (x - 4)$ . This is $F = - k (x - x_{0})$ with $k = 2 N m^{- 1}$ and $x_{0} = 4 m$ .

(a) Yes, the motion is SHM. (b) Equilibrium (mean) position: $x = 4 m$ .

(c) Released at rest from $x = 6 m$ , so this is an extreme: $A = 6 - 4 = 2 m$ , and $ω = k / m = 2/2 = 1 s^{- 1}$ . Starting at the positive extreme: $x - 4 = 2 cos t$ , that is $x = 4 + 2 cos t$ (SI units).

(d) $T = \frac{2 π}{ω} = 2 π s \approx 6.28 s$ .

Solved Example 7

A particle starts from

x = A /2

and moves towards the positive extreme. Find the equation of the SHM using the reference circle. The amplitude is

A

Solution:

Draw the reference circle of radius $A$ and a vertical line through $x = A /2$ ; it cuts the circle at two points, one in the upper half ( $R$ ) and one in the lower half ( $S$ ).
For clockwise motion, the point in the upper half has its horizontal velocity towards $+ A$ ; the lower one moves towards $O$ . So the particle starts at $R$ .
In the triangle formed by $O$ , $R$ and the foot of the perpendicular, $sin ϕ = \frac{A /2}{A}$ , so the angle from the upper vertical is $ϕ = 3 0^{\circ}$ .

Answer: $x = A sin (ω t + π /6)$ .

Solved Example 8

A particle starts from

x = - \frac{3}{2} A

and moves towards the negative extreme. (a) Find the equation of the SHM. (b) Find the time taken to go directly from its initial position to the negative extreme. (c) Find the time taken to reach the mean position.

Solution:

(a) $sin ϕ = - \frac{3}{2}$ and the velocity $A ω cos ϕ$ must be negative, so $ϕ$ lies in the third quadrant: $ϕ = π + \frac{π}{3} = \frac{4 π}{3}$ . $x = A sin (ω t + \frac{4 π}{3})$ .

(b) The negative extreme is phase $\frac{3 π}{2}$ . Angle to turn: $\frac{3 π}{2} - \frac{4 π}{3} = \frac{π}{6}$ . Time $t = \frac{π /6}{ω} = \frac{T}{12}$ .

(c) The particle first goes to $- A$ and then comes back to $O$ (phase $2 π$ ). Angle $= 2 π - \frac{4 π}{3} = \frac{2 π}{3}$ , so $t = \frac{2 π /3}{2 π} T = \frac{T}{3}$ .

Solved Example 9

Two particles undergo SHM along parallel lines with the same time period

T

and equal amplitudes. At a particular instant, one is at its extreme position while the other is at its mean position. They move in the same direction. They will cross each other after a further time
(A)

T /8

(B)

3 T /8

(C)

T /6

(D)

4 T /3

Solution:

Let $Q$ be at $+ A$ . From here $Q$ moves towards $- x$ , so "same direction" means $P$ , at the mean position, is also moving towards $- x$ . On the reference circle $Q$ is at $9 0^{\circ}$ and $P$ at $18 0^{\circ}$ : phase difference $π /2$ , which never changes.

Shadows coincide when the two radii are mirror images about the vertical diameter. With a fixed gap of $9 0^{\circ}$ this first happens at $Q$ : $22 5^{\circ}$ , $P$ : $31 5^{\circ}$ , both at $x = - A / 2$ . Each radius has turned $\frac{3 π}{4}$ , so

t = \frac{3 π /4}{2 π} T = \frac{3 T}{8}

Check: $x_{Q} = A cos ω t$ , $x_{P} = - A sin ω t$ are equal when $tan ω t = - 1$ , that is $ω t = 3 π /4$ .

Figure 14: Both radii turn through

3 π /4

; the shadows of

P^{'}

and

Q^{'}

coincide at

x = - A / 2

after

t = 3 T /8

Answer: (B).

Solved Example 10

Two particles execute SHM of the same amplitude

20 cm

with the same period along the same line about the same equilibrium position. If their phase difference is

π /3

, find the maximum distance between them.

Solution:

$x_{1} = A sin ω t$ , $x_{2} = A sin (ω t + π /3)$ . Their separation is

x_{2} - x_{1} = 2 A sin \frac{π}{6} cos (ω t + \frac{π}{6}) = A cos (ω t + \frac{π}{6})

an SHM of amplitude $2 A sin (π /6) = A$ . On the reference circle, the chord joining the two points subtends $6 0^{\circ}$ , so it equals $A$ ; the distance is largest when this chord is horizontal.

Answer: maximum distance $= A = 20 cm$ .

Solved Example 11

Two particles execute SHM of the same time period along the same line about the same mean position. One starts from the mean position (amplitude

A

, moving towards the other) and the other from its extreme position (amplitude

2 A

). When will they meet?

Solution:

$x_{1} = A sin ω t$ and $x_{2} = 2 A cos ω t$ (phase difference $π /2$ ). They meet when $A sin ω t = 2 A cos ω t$ , that is $tan ω t = 2$ .

On the reference circles (radii $A$ and $2 A$ ) this is the angle where the two shadows coincide: if the second radius makes angle $θ$ with the horizontal, $A cos θ = 2 A sin θ$ , so $θ = tan^{- 1} (1/2)$ and the first radius has turned $\frac{π}{2} - θ = tan^{- 1} 2$ .

Answer: $t = \frac{1}{ω} tan^{- 1} 2 \approx \frac{1.107}{ω} \approx 0.176 T$ .

Solved Example 12

Two particles have time periods

T

and

5 T /4

. They start SHM at the same time from the mean position. After how many oscillations of the particle with the smaller period will they be in the same phase again?

Solution:

They are again in the same phase when both are back at the mean position moving the same way, that is when each has completed a whole number of oscillations in the same time: $n_{1} T = n_{2} \frac{5 T}{4}$ , so $\frac{n _{1}}{n _{2}} = \frac{5}{4}$ . The smallest whole numbers are $n_{1} = 5$ , $n_{2} = 4$ .

Answer: after 5 oscillations of the faster particle (time $5 T$ ).

Solved Example 13

A particle of mass

0.50 kg

executes SHM under a force

F = - (50 N m^{- 1}) x

. If it crosses the centre of oscillation with a speed of

10 m s^{- 1}

, find the amplitude.

Solution:

At the centre, $U = 0$ and all the energy is kinetic: $E = \frac{1}{2} m v^{2} = \frac{1}{2} (0.50) (10)^{2} = 25 J$ .

At the extreme, $K = 0$ and $E = \frac{1}{2} k A^{2}$ with $k = 50 N m^{- 1}$ : $\frac{1}{2} (50) A^{2} = 25$ , so $A^{2} = 1$ .

Answer: $A = 1 m$ . Check: $ω = k / m = 10 s^{- 1}$ and $A = v_{m a x} / ω = 10/10 = 1 m$ .

Solved Example 14

A particle in SHM has speed

8 cm s^{- 1}

x = 3 cm

and

6 cm s^{- 1}

x = 4 cm

. Find

ω

, the amplitude and the time period.

Solution:

$v^{2} = ω^{2} (A^{2} - x^{2})$ at both points: $64 = ω^{2} (A^{2} - 9)$ and $36 = ω^{2} (A^{2} - 16)$ . Subtract: $28 = 7 ω^{2}$ , so $ω = 2 s^{- 1}$ . Then $A^{2} = 9 + 64/4 = 25$ .

Answer: $ω = 2 rad s^{- 1}$ , $A = 5 cm$ , $T = 2 π / ω = π s \approx 3.14 s$ .

Solved Example 15

In SHM of amplitude

A

, the kinetic energy is three times the potential energy at a displacement of
(A)

A /3

(B)

A /2

(C)

A / 3

(D)

3 A /2

Solution:

$K = 3 U$ means $E = K + U = 4 U$ : $\frac{1}{2} k A^{2} = 4 \times \frac{1}{2} k x^{2}$ , so $x^{2} = A^{2} /4$ .

Answer: (B). At $x = A /2$ , $U = E /4$ and $K = 3 E /4$ .

Solved Example 16

A particle of mass

m

moves along the

x

-axis in the potential

U (x) = U_{0} (1 - cos a x)

, where

U_{0}

and

a

are positive constants. Find the time period of small oscillations about

x = 0

Solution:

$U^{'} (x) = U_{0} a sin a x$ is zero at $x = 0$ , and $U^{''} (0) = U_{0} a^{2} > 0$ , so $x = 0$ is a stable equilibrium. For small $x$ , $F = - U^{'} (x) \approx - U_{0} a^{2} x$ , which is SHM with $k = U_{0} a^{2}$ .

ω = \frac{U ^{''} ( 0 )}{m} = a \frac{U _{0}}{m}, T = \frac{2 π}{a} \frac{m}{U _{0}}

Answer: $T = \frac{2 π}{a} m / U_{0}$ .

Practice Questions

Which of these forces give SHM: (i) $F = - 5 x$ (ii) $F = - 5 x^{3}$ (iii) $F = 10 - 5 x$ (iv) $F = 5 x - 10$ ?Answer: (i) and (iii). (ii) oscillates but is not SHM; (iv) pushes the particle away from $x = 2$ .
A particle in SHM has amplitude $4 cm$ and period $2 s$ . Find its speed at $x = 2 cm$ .Answer: $v = π 16 - 4 = 23 π \approx 10.9 cm s^{- 1}$ .
Find the minimum time for a particle in SHM of period $12 s$ to go from $x = A /2$ to $x = - A /2$ .Answer: $T /6 = 2 s$ ( $6 0^{\circ}$ on the reference circle).
The maximum speed and maximum acceleration of an SHM are $10 m s^{- 1}$ and $50 m s^{- 2}$ . Find $ω$ and $A$ .Answer: $ω = 50/10 = 5 s^{- 1}$ ; $A = 1 0^{2} /50 = 2 m$ .
At what displacement is the kinetic energy of an SHM $75%$ of its total energy?Answer: $U = E /4$ , so $x = \pm A /2$ .
Write the equation of an SHM of amplitude $3 cm$ and period $4 s$ that starts at $x = - 3 cm$ .Answer: $x = - 3 cos (\frac{π t}{2}) cm$ , that is $x = 3 sin (\frac{π t}{2} + \frac{3 π}{2}) cm$ .
Two identical SHMs differ in phase by $π /2$ . What is the maximum separation between the particles if $A = 5 cm$ ?Answer: $2 A sin (π /4) = 52 \approx 7.07 cm$ .

Common Mistakes to Avoid

Watch out

Taking the phase constant from the position alone. $x_{0} = A /2$ gives two angles ( $π /6$ and $5 π /6$ ); the direction of velocity decides.
Writing $ω = m / k$ or $T = 2 π k / m$ . Check: a heavier mass must oscillate more slowly.
Using $v = ω A^{2} - x^{2}$ with $x$ measured from the origin when the mean position is elsewhere. Always measure $x$ from the mean position.
Thinking acceleration and velocity are both zero at the extremes. At $x = \pm A$ the speed is zero but the acceleration is maximum.
Assuming equal distances take equal times. $O \to A /2$ takes $T /12$ but $A /2 \to A$ takes $T /6$ .
Forgetting that energy goes as $A^{2}$ : doubling the amplitude makes the energy four times, not two times.
Saying kinetic energy has the same frequency as the SHM. $K$ and $U$ repeat every $T /2$ , so their frequency is $2 f$ .
Mixing degrees and radians in $ω t$ . $ω$ is in $rad s^{- 1}$ ; convert $π /6$ to $3 0^{\circ}$ only at the end.

Frequently Asked Questions

What is simple harmonic motion in simple words?

Simple harmonic motion is a to-and-fro motion about a mean position in which the restoring force is proportional to the displacement and always points back to the mean position, $F = - k x$ . A block on a spring and a pendulum swinging through a small angle are common examples.

What is the difference between oscillatory motion and simple harmonic motion?

All simple harmonic motion is oscillatory, but not all oscillatory motion is simple harmonic. Oscillatory motion only needs a restoring force towards a stable equilibrium. SHM needs that force to be exactly proportional to the displacement. A force like minus k x cubed gives oscillation that is not SHM.

Is every periodic motion simple harmonic?

No. Periodic motion only has to repeat after a fixed time. The Earth going round the Sun is periodic but not to-and-fro, so it is not SHM. SHM is a special periodic motion with a sinusoidal displacement, described by x equal to A sin of omega t plus phi.

Why is acceleration maximum at the extreme position in SHM?

Acceleration equals minus omega squared times displacement. The displacement is largest at the extremes, so the restoring force and the acceleration are largest there, even though the particle is momentarily at rest. At the mean position the displacement, force and acceleration are all zero while speed is maximum.

How do you find the phase constant of an SHM?

Put t equal to zero in x equal to A sin phi to get two possible angles, then use the sign of the starting velocity, A omega cos phi, to choose one. On the reference circle, phi is the angle of the starting point measured clockwise from the upper vertical.

At what position are kinetic and potential energy equal in SHM?

Kinetic and potential energy are equal at a displacement of amplitude divided by root two, about 0.71 A, on either side of the mean position. There each is half of the total energy, which stays constant at one half k A squared throughout the motion.

Which SHM topics are most important for JEE Main?

JEE Main regularly tests phase and the reference circle method, time taken between two positions, velocity and acceleration at a given displacement, energy graphs, and time periods of spring and pendulum systems. Learn the time ladder T/12, T/8, T/6, T/4 and the relation v equals omega root of A squared minus x squared.

How is simple harmonic motion asked in NEET?

NEET mostly asks direct, formula-based questions from SHM: maximum speed A omega, maximum acceleration omega squared A, energy at a given displacement, phase difference between displacement, velocity and acceleration, and reading x-t or energy graphs. Clear formulas and graph shapes are usually enough to score full marks.

Previous year questions on Simple Harmonic Motion And Oscillation

27 questions from past papers, each with a step-by-step solution.

Show all 27 questions

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1. Periodic and Oscillatory Motion

1.1 Periodic motion

1.2 Oscillatory motion

2. The Oscillatory Equation F=−kxn

3. Types of SHM

4. Motion of the Particle in One Cycle

5. Characteristics of SHM

6. Equation of SHM

6.1 The differential equation

6.2 Finding ϕ from the starting point

7. Velocity and Acceleration in SHM

7.1 Velocity

7.2 Acceleration

8. Graphs of Displacement, Velocity and Acceleration

9. SHM as the Projection of Uniform Circular Motion

9.1 Problem-solving strategy (horizontal phasor)

10. Energy in SHM

10.1 Kinetic energy

10.2 Potential energy

10.3 Total mechanical energy

11. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is simple harmonic motion in simple words?

What is the difference between oscillatory motion and simple harmonic motion?

Is every periodic motion simple harmonic?

Why is acceleration maximum at the extreme position in SHM?

How do you find the phase constant of an SHM?

At what position are kinetic and potential energy equal in SHM?

Which SHM topics are most important for JEE Main?

How is simple harmonic motion asked in NEET?

Previous year questions on Simple Harmonic Motion And Oscillation

Ready to master Oscillations?

2. The Oscillatory Equation $F = - k x^{n}$

6.2 Finding $ϕ$ from the starting point