SHM in Pendulums

Q: What is the time period of a simple pendulum?

For small swings the time period of a simple pendulum is T = 2π√(ℓ/g), where ℓ is the length from the point of suspension to the centre of the bob. It does not depend on the mass of the bob or on the amplitude, as long as the swing is small.

PhysicsOscillationsFor NEET aspirants

A pendulum is a body that swings about a fixed point under gravity; for small swings its motion is angular simple harmonic motion. SHM in pendulums gives $T = 2 π ℓ / g$ for a simple pendulum, $T = 2 π I / m g d$ for a compound (physical) pendulum and $T = 2 π I / C$ for a torsional pendulum. This page derives each one, then covers pendulums in lifts and cars, clocks that gain or lose time, the seconds pendulum and the large-amplitude correction. Pendulum questions are regular in JEE Main and NEET.

On this page1Angular SHM2Simple pendulum3Accelerating frames4Clocks5Compound pendulum6Rods with springs7Torsional pendulum

Key Formulas - Quick Reference

★ Must learnAngular SHM: $τ = - k θ$ , $\frac{d ^{2} θ}{d t ^{2}} + ω^{2} θ = 0$ , $ω = k / I$
★ Must learnSimple pendulum (small $θ$ ): $T = 2 π \frac{ℓ}{g}$ (independent of mass and small amplitude)
Seconds pendulum: $T = 2 s$ , $ℓ = g / π^{2} \approx 0.99 m$
★ Must learnAccelerating support: $T = 2 π ℓ / g_{eff}$ , $g_{eff} = g - a$ ; lift up $g + a$ , lift down $g - a$ , car $g^{2} + a^{2}$
Small changes: $\frac{Δ T}{T} = \frac{1}{2} \frac{Δ ℓ}{ℓ} - \frac{1}{2} \frac{Δ g}{g}$ ; a clock with period $T^{'}$ instead of $T$ loses $\frac{T ^{'} - T}{T ^{'}} t$ in time $t$
★ Must learnCompound pendulum: $T = 2 π \frac{I}{m g d}$ ; equivalent length $L = \frac{I}{m d} = d + \frac{K ^{2}}{d}$
Minimum period of a compound pendulum when $d = K$ (radius of gyration about the centre of mass)
★ Must learnTorsional pendulum: $τ = - C θ$ , $T = 2 π I / C$
Large amplitude: $T \approx T_{0} (1 + \frac{θ _{0}^{2}}{16})$ , $θ_{0}$ in radians

1. Angular SHM

If a body turns about a fixed axis and the restoring torque is proportional to its angular displacement from the mean position, the motion is angular SHM:

τ = - k θ \Rightarrow I \frac{d ^{2} θ}{d t ^{2}} = - k θ \Rightarrow \frac{d ^{2} θ}{d t ^{2}} + ω^{2} θ = 0, ω = \frac{k}{I}

Here $k$ is the SHM (torsional) constant in $N m rad^{- 1}$ and $I$ is the moment of inertia about the axis. The solution is $θ = θ_{0} sin (ω t + ϕ)$ , with angular velocity $\frac{d θ}{d t} = θ_{0} ω cos (ω t + ϕ)$ and angular acceleration $- ω^{2} θ$ .

Linear SHM	Angular SHM
displacement $x$	angular displacement $θ$
mass $m$	moment of inertia $I$
force $F = - k x$	torque $τ = - k θ$
$T = 2 π m / k$	$T = 2 π I / k$
$E = \frac{1}{2} k A^{2}$	$E = \frac{1}{2} k θ_{0}^{2}$

Key idea

Every pendulum on this page is solved the same way: find the restoring torque, write it as

- k θ

, then

T = 2 π I / k

2. Simple Pendulum

A heavy point mass hung by a weightless, inextensible and perfectly flexible string from a rigid support is a simple pendulum. Its length $ℓ$ is measured from the point of suspension to the centre of the bob.

Figure 1: Only

m g sin θ

acts along the path, towards the mean position. For small

θ

m g sin θ \approx m g θ

, which makes the motion SHM.

2.1 Derivation (torque method)

Take torques about the point of suspension $O$ . The tension passes through $O$ , so $τ_{T} = 0$ .
The weight gives $τ = - m g ℓ sin θ$ (it acts to reduce $θ$ ).
For small $θ$ (in radians), $sin θ \approx θ$ : $τ = - m g ℓ θ$ . Compare with $τ = - k θ$ : $k = m g ℓ$ .
$I = m ℓ^{2}$ for a point bob, so
$T = 2 π \frac{I}{k} = 2 π \frac{m ℓ ^{2}}{m g ℓ} = 2 π \frac{ℓ}{g}$

Force view: along the arc the restoring force is $- m g sin θ \approx - m g \frac{s}{ℓ}$ , where $s = ℓ θ$ is the arc length. This is $F = - k s$ with $k = m g / ℓ$ , giving the same period. So a simple pendulum performs angular SHM, but for small swings it can be treated as linear SHM along the arc.

$T$ does not depend on the mass of the bob or on the amplitude (as long as it is small).
$T \propto ℓ$ and $T \propto 1/ g$ . Plotting $T^{2}$ against $ℓ$ gives a straight line of slope $4 π^{2} / g$ .
Seconds pendulum: period $2 s$ (each swing one second), length $ℓ = g / π^{2} \approx 0.99 m$ . Sometimes $g$ is taken as $π^{2}$ to simplify calculations, making $ℓ = 1 m$ .

Figure 2:

T \propto ℓ

curves (left);

T^{2} = \frac{4 π ^{2}}{g} ℓ

is a straight line (right). The slope of the

T^{2}

ℓ

line gives

g = 4 π^{2} / slope

Figure 3: Exact

T / T_{0}

(solid) and

1 + θ_{0}^{2} /16

(dashed). At

1 0^{\circ}

the error of

T_{0} = 2 π ℓ / g

is only 0.2%, so the small-angle formula is safe.

JEE Advanced

Beyond the small-angle formula. For amplitude $θ_{0}$ , $T \approx T_{0} (1 + \frac{θ _{0}^{2}}{16})$ . At $1 0^{\circ}$ the error is only 0.2%, at $9 0^{\circ}$ the true period is about $18%$ longer. Very long pendulum: if $ℓ$ is comparable with the Earth's radius $R$ , $T = 2 π \frac{1}{g ( 1/ ℓ + 1/ R )}$ ; even an infinitely long pendulum has a finite period $2 π R / g \approx 84.6 min$ .

3. Pendulum in an Accelerating Frame

If the point of suspension accelerates with $a$ , work in its frame: add a pseudo force $- m a$ to the bob. The bob then feels an effective gravity

g_{eff} = g - a, T = 2 π \frac{ℓ}{g _{eff}}

This holds when $∣ g - a ∣$ is constant. The mean position lies along $g_{eff}$ .

Figure 4: Replace

g

g_{eff} = ∣ g - a ∣

. Upward acceleration shortens the period, downward lengthens it; sideways acceleration tilts the mean position by

tan θ_{0} = a / g

Situation	$g_{eff}$	Effect on $T$
Lift accelerating up (or decelerating while going down)	$g + a$	decreases
Lift accelerating down (or decelerating while going up)	$g - a$	increases
Lift in free fall ( $a = g$ )	$0$	no oscillation ( $T \to \infty$ )
Car with horizontal acceleration $a$	$g^{2} + a^{2}$ ; string tilts by $tan θ_{0} = a / g$	decreases
Bob of density $ρ$ swinging in a liquid of density $σ$ (drag ignored)	$g (1 - \frac{σ}{ρ})$	increases
Uniform lift or train (constant velocity)	$g$	unchanged

Exam Trick

Draw $g_{eff}$ first. Put the pseudo acceleration $- a$ tail to head with $g$ ; the resultant gives both the new period and the direction of the string at rest. Any constant extra force works the same way: a charged bob in a vertical field $E$ has $g_{eff} = g \pm q E / m$ .

Figure 5: One rule,

T = 2 π ℓ / g_{eff}

. (a)

a = g /2

T^{'} = 0.946 T

, string tilted

26. 6^{\circ}

. (b)

q E = m g /3

T^{'} = 0.866 T

. (c)

ρ = 8 σ

, buoyancy

B = m g /8

T^{'} = 1.069 T

(Solved Example 10).

Key idea

Whatever the set-up, find the net constant force per unit mass on the bob at rest: that is

g_{eff}

, and

T = 2 π ℓ / g_{eff}

4. Pendulum Clocks: Gaining and Losing Time

A pendulum clock counts swings. If its period grows, it counts fewer swings in a given real time and runs slow (loses time); if the period shrinks, it runs fast (gains time).

Correct period $T$ , actual period $T^{'}$ . In real time $t$ the clock makes $t / T^{'}$ swings and shows $\frac{t}{T ^{'}} T$ .
Time lost $= t - \frac{t}{T ^{'}} T = \frac{T ^{'} - T}{T ^{'}} t$ (negative means time gained).
For small changes, $\frac{Δ T}{T} = \frac{1}{2} \frac{Δ ℓ}{ℓ} - \frac{1}{2} \frac{Δ g}{g}$ , and the time lost in a day $\approx \frac{Δ T}{T} \times 86400 s$ .

Exam Trick

Half the percentage. A $1%$ change in $ℓ$ or $g$ changes $T$ by $0.5%$ . Clock taken up a mountain ( $g$ less): slow. Pendulum rod heats and expands ( $ℓ$ more): slow. Clock at the poles ( $g$ more): fast.

Figure 6: The clock shows

\frac{T}{T ^{'}} t

. With

T^{'} = 3 s

instead of

2 s

it reads

40

min after one real hour: it loses

20

min (Solved Example 1). With

T^{'} = 1.6 s

it reads

75

min and gains

15

min. Longer period, slow clock.

Quick Recall: tap to check

A pendulum clock is taken to a hill station. Does it gain or lose time?

It loses time:

g

is smaller, so

T

is longer and the clock runs slow.

In a lift accelerating downwards at

a < g

, does the period increase or decrease?

It increases:

g_{eff} = g - a

The length of a pendulum rises by

2%

. What happens to its period?

It rises by about

1%

(half the percentage).

5. Compound (Physical) Pendulum

A rigid body of any shape that swings in a vertical plane about a horizontal axis through it is a compound or physical pendulum. Let $S$ be the point of suspension, $C$ the centre of mass and $d = S C$ (constant during the motion).

Figure 7: A rigid body pivoted at

S

. The weight at

C

gives torque

- m g d sin θ

, so

T = 2 π I / m g d

. It swings like a simple pendulum of length

L = I / m d

(point

O

Restoring torque of the weight about $S$ : $τ = - m g d sin θ \approx - m g d θ$ for small $θ$ .
$I α = - m g d θ$ , where $I$ is the moment of inertia about the axis through $S$ . So $α = - \frac{m g d}{I} θ$ and $ω^{2} = \frac{m g d}{I}$ .
Hence
$T = 2 π \frac{I}{m g d}$

★ Must learnEquivalent length. Writing

I = m (K^{2} + d^{2})

(parallel-axis theorem,

K

= radius of gyration about the centre of mass):

T = 2 π \frac{L}{g}, L = \frac{I}{m d} = d + \frac{K ^{2}}{d}

The body swings like a simple pendulum of length

L

. The point

O

at distance

L

from

S

is the centre of oscillation; suspending the body from

O

gives the same period.

L

, and so

T

, is least when

d = K

Body and pivot	$I$ about pivot	$d$	Equivalent length $L$
Uniform rod of length $ℓ$ , pivot at one end	$\frac{m ℓ ^{2}}{3}$	$\frac{ℓ}{2}$	$\frac{2 ℓ}{3}$
Ring of radius $R$ , pivot on the rim	$2 m R^{2}$	$R$	$2 R$
Disc of radius $R$ , pivot on the rim	$\frac{3}{2} m R^{2}$	$R$	$\frac{3 R}{2}$
Disc, pivot at distance $z$ from the centre	$\frac{m R ^{2}}{2} + m z^{2}$	$z$	$z + \frac{R ^{2}}{2 z}$

Figure 8: Parallel-axis theorem gives

I

about the pivot:

2 M R^{2}

for the ring,

\frac{1}{2} M R^{2} + M z^{2}

for the disc. Then

T = 2 π I / M g d

Figure 9:

T \propto \frac{R ^{2}}{2 z} + z

. The period is least when

z = R / 2

, the radius of gyration of the disc about its centre.

Simple pendulum

Point mass on a massless string; all the mass is at distance $ℓ$ . $I = m ℓ^{2}$ , $d = ℓ$ , so $T = 2 π ℓ / g$ .

Compound pendulum

Rigid body; mass is spread out. $I = m (K^{2} + d^{2})$ , so it swings like a simple pendulum of length $L = d + K^{2} / d > d$ .

Quick Recall: tap to check

A simple pendulum and a uniform rod pivoted at one end have the same length. Which has the shorter period?

The rod: its equivalent length is

2 ℓ /3

Where should a disc be pivoted for the smallest period?

z = R / 2

from its centre, where

d

equals the radius of gyration.

Does the mass of a compound pendulum affect its period?

No:

I \propto m

, so

m

cancels in

I / m g d

6. Rods and Bodies Held by Springs

When springs act on a pivoted body, add their torques to the gravitational torque. For a small turn $θ$ , a spring attached at distance $r$ from the pivot is deformed by $r θ$ and gives a torque $- (k r θ) r = - k r^{2} θ$ .

Figure 10: Both springs and gravity give restoring torques:

τ = - (2 k ℓ^{2} + m g ℓ /2) θ

. With

I = m ℓ^{2} /3

ω = \frac{3 ( 4 k ℓ + m g )}{2 m ℓ}

Exam Trick

Add the stiffnesses as torques. $k_{total} = \sum k_{i} r_{i}^{2} \pm m g d$ (plus if the centre of mass hangs below the pivot, minus if it is above it). Then $ω = k_{total} / I$ . If the result is negative, the equilibrium is unstable and there is no SHM.

7. Torsional Pendulum

An extended body hung at its centre from a torsion wire forms a torsional pendulum. The wire does not stretch but twists about its axis. Turn the body by a small angle $θ$ and release it: the twisted wire applies a restoring torque

τ = - C θ \Rightarrow I α = - C θ \Rightarrow α = - \frac{C}{I} θ, T = 2 π \frac{I}{C}

$C$ is the torsional constant of the wire ( $N m rad^{- 1}$ ) and $I$ is the moment of inertia about the wire. Gravity plays no part, so the period is the same anywhere. The oscillation stays simple harmonic even for fairly large twists, as long as the wire obeys Hooke's law.

Figure 11: Twisting the wire by

θ

produces a restoring torque

τ = - C θ

, so the disc performs angular SHM with

T = 2 π I / C

Inertia table. The torsional pendulum is used to measure an unknown moment of inertia: measure the period with a known body and with the unknown body on the same wire; then $\frac{I _{2}}{I _{1}} = \frac{T _{2}^{2}}{T _{1}^{2}}$ .

JEE Advanced

For a wire of length $ℓ$ , radius $r$ and modulus of rigidity $G$ , $C = \frac{π G r ^{4}}{2 ℓ}$ . The period is very sensitive to the radius ( $C \propto r^{4}$ ), which is why galvanometers and Cavendish balances use very fine wires.

Key idea

Only the torsional pendulum ignores gravity: its restoring torque comes from the twisted wire, so

T = 2 π I / C

is the same on the Moon.

8. All Pendulums at a Glance

Pendulum	Restoring torque	Period	Depends on $g$ ?
Simple	$- m g ℓ θ$	$2 π ℓ / g$	yes
Compound (physical)	$- m g d θ$	$2 π I / m g d$	yes
Torsional	$- C θ$	$2 π I / C$	no
Pivoted body with springs	$- (\sum k r^{2} \pm m g d) θ$	$2 π I / k_{total}$	partly

Figure 12: Every pendulum question is one torque:

τ = - k θ

and

T = 2 π I / k

. The two notes cover the favourite twists:

g_{eff}

for lifts, cars and charged bobs, and extra spring torques.

Quick Recall: tap to check

Which pendulum's period does not depend on

g

The torsional pendulum:

T = 2 π I / C

A uniform rod pivoted at one end: what is its equivalent length?

L = 2 ℓ /3

A charged bob (

+ q

) swings in a downward field

E

. What is

g_{eff}

g + q E / m

, so the period decreases.

Figure 13: All pendulums on one screen. Each branch is the same recipe: restoring torque

- k θ

, then

T = 2 π I / k

9. Solved Examples

Solved Example 1

The period of a pendulum clock that should be

2 s

becomes

3 s

. How much time does the clock lose in one hour?

Solution:

Each real $3 s$ the clock completes one swing and shows $2 s$ , so it loses $1 s$ every $3 s$ , that is $\frac{1}{3} s$ per second.

Time lost $= \frac{T ^{'} - T}{T ^{'}} t = \frac{1}{3} \times 3600$ . Answer: $1200 s$ (20 minutes) per hour.

Solved Example 2

A simple pendulum of length

ℓ

hangs from the ceiling of a car accelerating uniformly with

a_{0}

on a horizontal road. Find the period of small oscillations about the mean position.

Solution:

In the car frame a pseudo force $m a_{0}$ acts backwards on the bob. At the mean position the tension balances the resultant of $m g$ and $m a_{0}$ , $F = m g^{2} + a_{0}^{2}$ , so the string makes $θ_{0}$ with the vertical, $tan θ_{0} = \frac{a _{0}}{g}$ .

Deflect the string by a further small $θ$ : restoring torque $= - (F sin θ) ℓ \approx - m g^{2} + a_{0}^{2} ℓ θ$ and $I = m ℓ^{2}$ , so $α = - \frac{g ^{2} + a _{0}^{2}}{ℓ} θ$ .

Answer:

T = 2 π \frac{ℓ}{g ^{2} + a _{0}^{2}} = \frac{2 π ℓ}{( g ^{2} + a _{0}^{2} ) ^{1/4}}

Solved Example 3

A ring hung on a nail at a point on its rim oscillates in its own plane as a seconds pendulum. Find its radius. Take

g = π^{2} m s^{- 2}

Solution:

Seconds pendulum: $T = 2 s$ . About the nail, $I = M R^{2} + M R^{2} = 2 M R^{2}$ and $d = R$ .

$2 = 2 π \frac{2 M R ^{2}}{M g R} = 2 π \frac{2 R}{π ^{2}}$ , so $\frac{2 R}{π ^{2}} = \frac{1}{π ^{2}}$ .

Answer: $R = 0.5 m$ .

Solved Example 4

A circular disc of mass

M

and radius

R

has a tiny hole at distance

z

from its centre (

z < R

). A horizontal shaft through the hole lets it swing in a vertical plane. For what

z

is the period of small oscillations minimum?

Solution:

$I = \frac{M R ^{2}}{2} + M z^{2}$ , $d = z$ , so

T = 2 π \frac{R ^{2} /2 + z ^{2}}{g z} = 2 π \frac{1}{g} (\frac{R ^{2}}{2 z} + z)

Minimise $f (z) = \frac{R ^{2}}{2 z} + z$ : $\frac{df}{d z} = - \frac{R ^{2}}{2 z ^{2}} + 1 = 0$ .

Answer: $z = \frac{R}{2}$ (Figure 9), the radius of gyration of the disc about its centre.

Solved Example 5

A uniform rod of mass

m

and length

ℓ

is hinged at its upper end. Its lower end is joined to two horizontal springs, each of constant

k

, fixed to opposite walls (both springs at natural length when the rod is vertical). Find the angular frequency of small oscillations.

Solution:

For a small turn $θ$ , the lower end moves $ℓ θ$ : one spring is compressed and the other stretched by $ℓ θ$ , each pushing back with $k ℓ θ$ at lever arm $ℓ$ . Gravity adds $- m g \frac{ℓ}{2} sin θ \approx - m g \frac{ℓ}{2} θ$ .

τ = - (2 k ℓ^{2} + \frac{m g ℓ}{2}) θ, I = \frac{m ℓ ^{2}}{3}

Answer:

ω = \frac{2 k ℓ ^{2} + m g ℓ /2}{m ℓ ^{2} /3} = \frac{3 ( 4 k ℓ + m g )}{2 m ℓ}

Solved Example 6

A uniform disc of radius

5.0 cm

and mass

200 g

is fixed at its centre to a metal wire whose other end is fixed to the ceiling. The disc is twisted and released; it makes torsional oscillations of period

0.20 s

. Find the torsional constant of the wire.

Solution:

$I = \frac{m r ^{2}}{2} = \frac{( 0.200 ) ( 5.0 \times 1 0 ^{- 2} ) ^{2}}{2} = 2.5 \times 1 0^{- 4} kg m^{2}$ .

$T = 2 π I / C$ , so $C = \frac{4 π ^{2} I}{T ^{2}} = \frac{4 π ^{2} ( 2.5 \times 1 0 ^{- 4} )}{( 0.20 ) ^{2}}$ .

Answer: $C \approx 0.25 N m rad^{- 1}$ (that is, $kg m^{2} s^{- 2}$ ).

Solved Example 7

Find the length of a seconds pendulum where

g = 9.8 m s^{- 2}

Solution:

$T = 2 s = 2 π ℓ / g$ , so $ℓ = \frac{g T ^{2}}{4 π ^{2}} = \frac{9.8}{π ^{2}}$ .

Answer: $ℓ \approx 0.993 m$ .

Solved Example 8

A simple pendulum has period

T

in a stationary lift. The lift now accelerates upwards at

g /3

. The new period is
(A)

T 3 /2

(B)

T 2/3

(C)

2 T / 3

(D)

T

Solution:

$g_{eff} = g + \frac{g}{3} = \frac{4 g}{3}$ , so $T^{'} = T \frac{g}{4 g /3} = T \frac{3}{4} = \frac{3}{2} T \approx 0.87 T$ .

Answer: (A).

Solved Example 9

A pendulum clock keeps correct time at sea level. It is taken to a mountain top where

g

0.2%

smaller. How much time does it lose per day?

Solution:

$\frac{Δ T}{T} = - \frac{1}{2} \frac{Δ g}{g} = - \frac{1}{2} (- 0.2%) = + 0.1%$ : the period grows, so the clock runs slow.

Time lost per day $\approx 0.001 \times 86400$ . Answer: about $86 s$ per day.

Solved Example 10

A simple pendulum has a solid bob of density

8000 kg m^{- 3}

. It is made to swing with the bob completely inside water (density

1000 kg m^{- 3}

). Ignoring drag, by what factor does its period change?

Solution:

Buoyancy reduces the effective weight: $m g_{eff} = m g - \frac{m}{ρ} σ g$ , so $g_{eff} = g (1 - \frac{σ}{ρ}) = \frac{7 g}{8}$ .

Answer: $T^{'} = T 8/7 \approx 1.07 T$ (about $7%$ longer).

Solved Example 11

A uniform rod of length

ℓ

is pivoted at one end and oscillates in a vertical plane. Find its period and the length of the equivalent simple pendulum.

Solution:

$I = \frac{m ℓ ^{2}}{3}$ about the end, $d = \frac{ℓ}{2}$ . $T = 2 π \frac{m ℓ ^{2} /3}{m g ℓ /2} = 2 π \frac{2 ℓ}{3 g}$ .

Answer: $T = 2 π 2 ℓ /3 g$ ; equivalent length $L = 2 ℓ /3$ .

Solved Example 12

A simple pendulum of period

T

hangs from the roof of a car that accelerates horizontally with

a = 3 g

. At equilibrium the string makes an angle

θ_{0}

with the vertical, and the period of small oscillations is

T^{'}

. Then
(A)

θ_{0} = 3 0^{\circ}

T^{'} = T / 2

(B)

θ_{0} = 6 0^{\circ}

T^{'} = T / 2

(C)

θ_{0} = 6 0^{\circ}

T^{'} = T /2

(D)

θ_{0} = 3 0^{\circ}

T^{'} = 2 T

Solution:

$tan θ_{0} = \frac{a}{g} = 3$ , so $θ_{0} = 6 0^{\circ}$ . The effective gravity is $g_{eff} = g^{2} + a^{2} = g^{2} + 3 g^{2} = 2 g$ , so $T^{'} = T \frac{g}{2 g} = \frac{T}{2} \approx 0.71 T$ .

Answer: (B). Option (C) forgets the square root; (A) swaps $tan θ_{0} = g / a$ .

Practice Questions

A seconds pendulum on Earth is taken to the Moon ( $g_{Moon} = g /6$ ). Find its period there.Answer: $26 \approx 4.9 s$ .
The length of a simple pendulum is increased by $21%$ . By what percentage does its period increase?Answer: $1.21 = 1.1$ , so by $10%$ .
What is the period of a simple pendulum in a freely falling lift?Answer: Infinite: $g_{eff} = 0$ , so it does not oscillate.
A uniform rod $1 m$ long is pivoted at one end. Find its period ( $g = π^{2} m s^{- 2}$ ).Answer: $2 2/3 \approx 1.63 s$ .
Where should a uniform rod of length $ℓ$ be pivoted for the smallest period of oscillation?Answer: At $d = ℓ / 12 = ℓ / (23)$ from its centre.
A disc on a torsion wire has period $2 s$ . A ring with the same moment of inertia is placed on it. Find the new period.Answer: $I$ doubles, so $T = 22 \approx 2.83 s$ .
A pendulum hangs in a car accelerating horizontally at $a = g$ . Find the tilt of the string at rest and the new period in terms of $T$ .Answer: $4 5^{\circ}$ ; $T^{'} = T / 2^{1/4} \approx 0.84 T$ .

Common Mistakes to Avoid

Watch out

Using $T = 2 π ℓ / g$ for large swings. It needs small $θ$ (in practice below about $1 0^{\circ}$ ).
Measuring the length to the top of the bob. $ℓ$ runs from the point of suspension to the centre of the bob.
Thinking a heavier bob swings faster. The mass cancels; only $ℓ$ and $g$ matter.
Taking $g_{eff} = g - a$ for a lift accelerating upwards. Upward acceleration increases $g_{eff}$ to $g + a$ .
Using $I$ about the centre of mass in $T = 2 π I / m g d$ . $I$ must be about the pivot (parallel-axis theorem).
Saying a clock with a longer period gains time. A longer period means fewer ticks: the clock loses time.
Expecting gravity to change a torsional pendulum's period. $T = 2 π I / C$ has no $g$ .
Forgetting to convert degrees to radians in $sin θ \approx θ$ or in $1 + θ_{0}^{2} /16$ .

Frequently Asked Questions

What is the time period of a simple pendulum?

For small swings the time period of a simple pendulum is $T = 2 π ℓ / g$ , where $ℓ$ is the length from the point of suspension to the centre of the bob. It does not depend on the mass of the bob or on the amplitude, as long as the swing is small.

Why is a simple pendulum SHM only for small angles?

The restoring torque is proportional to sin theta, not to theta. Only for small angles is sin theta nearly equal to theta, making the torque proportional to the displacement. At larger amplitudes the motion is still periodic but not simple harmonic, and the period becomes slightly longer.

What is a seconds pendulum and what is its length?

A seconds pendulum has a period of 2 seconds, so each swing from one side to the other takes one second. With g equal to 9.8 metres per second squared its length is about 0.99 metre, which is why it is often rounded to 1 metre.

How does the period of a pendulum change in an accelerating lift?

Replace g by the effective gravity. In a lift accelerating upwards it becomes g plus a and the period decreases; accelerating downwards it becomes g minus a and the period increases. In free fall the effective gravity is zero and the pendulum does not oscillate at all.

What is the difference between a simple pendulum and a compound pendulum?

A simple pendulum is an idealised point mass on a massless string. A compound or physical pendulum is a real rigid body swinging about a pivot, with period two pi root of I over m g d. It behaves like a simple pendulum of equivalent length I over m d.

Does a torsional pendulum depend on gravity?

No. A torsional pendulum is restored by the twist of its wire, $τ = - C θ$ , so its period $T = 2 π I / C$ contains no $g$ . It would have the same period on the Moon. That is also why it is used to measure moments of inertia.

Which pendulum topics are important for JEE Main and JEE Advanced?

JEE tests pendulums in accelerating frames, clocks gaining or losing time, compound pendulums using the parallel-axis theorem, the minimum-period pivot point, rods held by springs, and torsional pendulums. JEE Advanced sometimes adds the large-amplitude correction or pendulums with charged bobs in electric fields.

What pendulum questions are asked in NEET?

NEET usually asks the simple pendulum formula, the effect of changing length or g, the seconds pendulum, pendulums in lifts, percentage change in period, and the energy of a swinging bob. Remember that a 1 percent change in length or g changes the period by about half a percent.

Previous year questions on SHM in Pendulums

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

Show all 12 questions

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