When will the motion of a simple pendulum be simple harmonic?

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Q. When will the motion of a simple pendulum be simple harmonic?

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Concept understanding — Simple Pendulum Period

Simple Pendulum Period: From Intuition to Formula

Imagine tying a small weight to a string, holding the other end fixed, and giving it a gentle push. It swings back and forth — that’s a simple pendulum. The question is: what determines how fast it swings? Does a heavier bob swing faster? Does a longer string make it slower?

Let’s start with what you already feel. If you hold a short string (say 20 cm) and swing it, the bob zips back and forth quickly. If you use a long string (say 1 m), the swing is noticeably slower. So length matters — longer means slower.

What about the weight? Try a light plastic bob and a heavy metal one of the same size, on the same string. You’ll find they swing at the same speed. That’s surprising — heavier things don’t fall faster, and here they don’t swing faster either. So mass does not affect the period (the time for one complete back-and-forth swing).

What about how hard you push? If you give a big push, the bob swings wider, but does it take more time? For small swings (small angles, say less than about 15°), the period is almost the same regardless of amplitude. That’s the key: for small oscillations, the pendulum is isochronous — its period is independent of amplitude.

Note

This is only true for small angles. If you pull the bob to 60° and let go, the period becomes noticeably longer. In most exam problems, you assume “small oscillations” (usually < 10°).

The Precise Statement

For a simple pendulum of length $L$ (measured from pivot to centre of bob), swinging with small amplitude in a uniform gravitational field $g$ , the time period $T$ (time for one complete oscillation) is:

T = 2\pi \sqrt{\frac{L}{g}}

That’s it. No mass term. No amplitude term (for small angles).

T = 2\pi \sqrt{\frac{L}{g}}

Why does this formula make sense?

$L$ in numerator: longer string → larger $T$ (slower swing). Doubling $L$ multiplies $T$ by $\sqrt{2} \approx 1.4$ .
$g$ in denominator: stronger gravity (larger $g$ ) → smaller $T$ (faster swing). On the Moon ( $g \approx 1.6$ m/s²), the same pendulum swings much slower.
$2\pi$ : comes from the mathematics of simple harmonic motion — the pendulum’s motion is approximately sinusoidal for small angles.

Tip

To remember: the formula is identical to that of a mass on a spring ( $T = 2\pi\sqrt{m/k}$ ), but here the “restoring force per unit displacement” is $mg/L$ , so the effective “ $k$ ” is $mg/L$ , giving $T = 2\pi\sqrt{L/g}$ .

Common exam pitfalls

Don’t confuse $L$ with amplitude. $L$ is the string length, not how far you pull it. …

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A simple pendulum executes simple harmonic motion (SHM) only when its angular displacement is small (typically $\theta \lesssim 10^\circ$ or $0.17$ rad), because only then does the restoring torque become directly proportional to the displacement. The period is then $T = 2\pi\sqrt{L/g}$ .

Why the pendulum is not always simple harmonic

A simple pendulum consists of a point mass $m$ attached to a massless, inextensible string of length $L$ , swinging under gravity. The restoring force that pulls the bob back toward the equilibrium position comes from the tangential component of gravity.

When the bob is displaced by an angle $\theta$ from the vertical, the gravitational force $mg$ splits into two components:

Radial: $mg\cos\theta$ (tension balances this)
Tangential: $mg\sin\theta$ (this is the restoring force)

The tangential force is $F = -mg\sin\theta$ , where the minus sign indicates it always points opposite to the displacement. For the motion to be simple harmonic, the restoring force must be directly proportional to the displacement — that is, $F \propto -\theta$ (or $F \propto -x$ for linear displacement).

Here lies the catch: $\sin\theta$ is not proportional to $\theta$ for large angles.

Watch out

Many students mistakenly write $F = -mg\theta$ directly. This is only valid when $\theta$ is small enough that $\sin\theta \approx \theta$ (in radians). For $\theta = 30^\circ$ ( $0.52$ rad), $\sin\theta = 0.5$ while $\theta = 0.52$ — a 4% error that grows rapidly with larger angles.

The small-angle approximation

For small angles measured in radians, the Taylor expansion of $\sin\theta$ gives:

\sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots

When $\theta \ll 1$ radian, the higher-order terms become negligible, and we can write:

\sin\theta \approx \theta

This approximation is excellent for $\theta < 0.17$ rad (about $10^\circ$ ), where the error is less than 0.5%.

F \approx -mg\theta \quad \text{for small } \theta

Since the arc length $s = L\theta$ , the linear displacement $x \approx s = L\theta$ (for small angles, the arc is nearly straight), so $\theta = x/L$ . Substituting:

F \approx -mg\left(\frac{x}{L}\right) = -\left(\frac{mg}{L}\right)x

This is exactly Hooke's law: $F = -kx$ with effective spring constant $k = mg/L$ . The motion is therefore simple harmonic.

Deriving the period

Using Newton's second law for rotational motion, the torque about the pivot is:

\tau = -mgL\sin\theta

For small $\theta$ , $\tau \approx -mgL\theta$ . Since $\tau = I\alpha$ and the moment of inertia of a point mass at distance $L$ is $I = mL^2$ :

mL^2\frac{d^2\theta}{dt^2} = -mgL\theta

\frac{d^2\theta}{dt^2} = -\frac{g}{L}\theta

This is the SHM equation $\frac{d^2\theta}{dt^2} = -\omega^2\theta$ , where $\omega = \sqrt{g/L}$ .

Tip

You don't need to re-derive this every time. The angular frequency $\omega = \sqrt{g/L}$ is the key result — memorize it, and the period follows directly. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$\dfrac{E}{8}$
›Reveal solutionSolution
Maximum KE in SHM is $E=\frac12 m\omega^2A^2$ , and for a simple pendulum $\omega^2=g/l$ . Doubling $l$ halves $\omega^2$ ; halving $A$ quarters $A^2$ — combined, the new max KE is $E/8$ .

Concept and Intuition

At the equilibrium (lowest) point of a pendulum's swing, all its energy is kinetic, and this maximum KE equals the total mechanical energy of the oscillation, $\frac12 m\omega^2A^2$ . For a simple pendulum, angular frequency depends only on length: $\omega=\sqrt{g/l}$ , so a longer pendulum swings more slowly per cycle. Both the change in $\omega$ (through $l$ ) and the change in amplitude directly shrink the peak kinetic energy, and their effects multiply.

Step-by-Step Solution

Maximum KE in SHM: $E_{max} = \tfrac12 m\omega^2 A^2$ .

For a simple pendulum: $\omega^2 = \dfrac{g}{l}$ , so $E = \tfrac12 m\dfrac{g}{l}A^2$ .

New length $l' = 2l$ : $\omega'^2 = \dfrac{g}{2l} = \dfrac{\omega^2}{2}$ .

New amplitude $A' = A/2$ : $A'^2 = \dfrac{A^2}{4}$ . …
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AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$\sigma$
›Reveal solutionSolution
Buoyancy in the liquid reduces the effective gravitational acceleration acting on the bob, giving $T=2\pi\sqrt{l/[g(1-\rho/\sigma)]}$ .

Concept and Intuition

A pendulum's period depends on the net downward acceleration of its bob. In air this is just $g$ ; in a liquid, buoyant upthrust partially cancels gravity. The apparent weight becomes $mg - V\rho g = mg\left(1-\dfrac{\rho}{\sigma}\right)$ (writing $V=m/\sigma$ ), so the effective acceleration is $g'=g(1-\rho/\sigma)$ .

Step-by-Step Solution

Weight of bob: $mg$ . Volume of bob: $V=m/\sigma$ .

Buoyant force: $F_b = V\rho g = \dfrac{m\rho g}{\sigma}$ .

Net downward force: $mg - \dfrac{m\rho g}{\sigma} = mg\left(1-\dfrac{\rho}{\sigma}\right)$ .

Effective $g' = g\left(1-\dfrac{\rho}{\sigma}\right)$ . …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$\dfrac{T}{t}$
›Reveal solutionSolution
Immersing a pendulum bob in a fluid reduces its effective weight via buoyancy, which lowers the effective $g$ and hence changes the period; the ratio of periods reduces to $\sqrt{g'/g}$ .

Concept and Intuition

For a simple pendulum, $T=2\pi\sqrt{L/g}$ . When the bob is submerged in a fluid, buoyancy acts upward, reducing the net restoring force per unit mass, which is equivalent to replacing $g$ with an effective $g'<g$ (for a bob denser than the fluid). The bob still executes SHM, just with this reduced effective gravity.

Step-by-Step Solution

Effective weight in fluid: $mg'=mg-\sigma Vg$ (buoyant force), where $V$ is bob volume and $\sigma$ is fluid density.

Since $m=\rho V$ (bob density $\rho$ ), dividing through by $m$ : $g'=g\left(1-\dfrac{\sigma}{\rho}\right)$ . …
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AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$\sqrt{1.732\,gR}$
›Reveal solutionSolution
The car's circular motion adds a horizontal pseudo-acceleration to gravity, making the effective gravity $2g$ while moving; at rest it drops back to $g$ , so the period increases by $\sqrt2$ .

Concept and Intuition

A simple pendulum's period depends on the effective gravitational field it swings in: $T=2\pi\sqrt{L/g_{eff}}$ . In a car going round a circular track, an observer inside the car (non-inertial frame) must add a centrifugal pseudo-acceleration $a_c=v^2/R$ , directed horizontally, to the real gravity $g$ , directed vertically. These two combine as perpendicular vectors.

Step-by-Step Solution

Given speed $v=\sqrt{1.732\,gR}$ , so $v^2=\sqrt3\,gR$ and $a_c=v^2/R=\sqrt3\,g$ .

The effective gravity (vector sum of vertical $g$ and horizontal $a_c$ , perpendicular to each other) is

$g_{eff}=\sqrt{g^2+(\sqrt3g)^2}=\sqrt{g^2+3g^2}=\sqrt{4g^2}=2g.$

While moving: $T=2\pi\sqrt{L/(2g)}$ .

At rest: $T_{rest}=2\pi\sqrt{L/g}$ . …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$20 \, \mathrm{ms^{-1}}$
›Reveal solutionSolution
The planet's gravity is first found from the projectile's total flight time (2s, symmetric), giving $g_p=20\ \mathrm{ms^{-2}}$ , which then gives a pendulum period of exactly 2 s.

Concept and Intuition

For a body thrown vertically upward and returning to the same launch point, the time to rise equals the time to fall (by symmetry of uniformly accelerated motion under gravity), so each takes half the total flight time. Once we know the planet's surface gravity $g_p$ , we can find the period of a simple pendulum there using $T = 2\pi\sqrt{L/g_p}$ .

Step-by-Step Solution

Total flight time = 2 s (thrown up, returns to same surface), so time to reach the top $t_{up} = 1$ s.

At the highest point, velocity is zero: $0 = u - g_p t_{up} \Rightarrow g_p = \dfrac{u}{t_{up}} = \dfrac{20}{1} = 20\ \mathrm{ms^{-2}}$ .

Pendulum length: $L = \dfrac{20}{\pi^2}$ m. …
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AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$x\ (> 1\text{m})$
›Reveal solutionSolution
The two pendulums realign in phase when the shorter one has completed exactly one extra oscillation; solving gives the longer pendulum's length as $9\ \mathrm{m}$ . Answer: (C).

Concept and Intuition

Two oscillators that start in phase will again be simultaneously at the mean position moving the same way whenever the faster one has gained a whole number of extra complete oscillations over the slower one. Since a pendulum's period depends on its length as $T\propto\sqrt{L}$ , this timing condition converts directly into a length ratio.

Step-by-Step Solution

Let $T$ be the period of the shorter (1 m) pendulum and $T_2$ the period of the longer ( $x$ m) pendulum, with $T_2 > T$ since $x>1$ .

At time $t$ , the shorter pendulum has completed $t/T$ oscillations and the longer one $t/T_2$ ; they are in phase again when $\dfrac{t}{T} - \dfrac{t}{T_2} = k$ for some positive integer $k$ .

Given $t = \dfrac{3T}{2}$ (the first such recurrence, so $k=1$ ): $\dfrac{3}{2} - \dfrac{3T}{2T_2} = 1$ . …
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AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
$Q. A pendulum is oscillating at a frequency of 8 Hz. Suddenly the string of the pendulum is clamped at its midpoint then the new frequency of oscillations is (A) 16 Hz (B) 13.8 Hz (C) 11.28 Hz (D) 5.7 Hz$
›Reveal solutionSolution
Clamping the string at its midpoint halves the effective pendulum length, so the new frequency

is $f\sqrt2 \approx 11.28$ Hz.

Concept and Intuition

A simple pendulum's frequency depends only on its length (and $g$ ): shorter pendulums swing faster.

When the string is clamped exactly at the midpoint, the point of support effectively moves down to

the clamp, and the bob now swings about that point with half the original length. Since

$f\propto 1/\sqrt L$ , cutting the length in half doesn't halve the frequency — it raises it by a

factor of $\sqrt2$ .

Step-by-Step Solution

Original relation: $f = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}=8\ \mathrm{Hz}$ .

New effective length: $L' = L/2$ .

New frequency: $f' = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{L/2}} = \dfrac{1}{2\pi}\sqrt{\dfrac{2g}{L}} = f\sqrt2$ .

Numerically: $f' = 8\times\sqrt2 \approx 8\times1.41 = 11.28\ \mathrm{Hz}$ . …
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AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
$\pi^2\ \mathrm{m\,s^{-2}}$
›Reveal solutionSolution
Rearranging $T=2\pi\sqrt{L/g}$ for $L$ with the convenient value $g=\pi^2\ \mathrm{m\,s^{-2}}$

gives $L=4$ m — the $\pi^2$ 's cancel neatly.

Concept and Intuition

The problem deliberately hands you $g=\pi^2\ \mathrm{m\,s^{-2}}$ so that the $\pi^2$ in the pendulum

formula cancels exactly, leaving a clean integer answer — a common trick to test whether you can

manipulate the formula rather than plug in $9.8$ .

Step-by-Step Solution

Start from $T = 2\pi\sqrt{\dfrac{L}{g}}$ .

Square both sides: $T^2 = 4\pi^2\dfrac{L}{g}$ . …
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AP EAPCET 2021Set ap-2021-09-07-FN1 markMCQ
$1\ m$
›Reveal solutionSolution
A seconds pendulum's length scales directly with local $g$ (since its period is fixed at $2\,s$ by definition); with lunar gravity $\approx g_{earth}/6$ , its length on the moon must be $1/6\ m$ .

Concept and Intuition

A 'seconds pendulum' is, by definition, a pendulum whose time period is exactly $2$ seconds (one full swing back and forth takes $2\ s$ ). The period of a simple pendulum is $T = 2\pi\sqrt{L/g}$ . If we require $T$ to stay fixed at $2\ s$ wherever the pendulum is used, then $L/g$ must stay constant, i.e. $L \propto g$ . Since the Moon's gravity is about $1/6$ of Earth's, a pendulum that keeps time as a 'seconds pendulum' on the Moon must be $1/6$ as long as the one on Earth.

Step-by-Step Solution

On Earth: $T = 2\pi\sqrt{\dfrac{L_E}{g_E}} = 2\ s$ , with $L_E = 1\ m$ .

On the Moon, for the SAME period $T=2\ s$ : $T = 2\pi\sqrt{\dfrac{L_M}{g_M}} = 2\pi\sqrt{\dfrac{L_E}{g_E}}$ . …
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AP EAPCET 2021Set ap-2021-10-05-FN1 markMCQ
$Q. If the length and time period of an oscillating pendulum have errors of 1% and 3% respectively, then the error in measurement of acceleration due to gravity is ______ (A) 4% (B) 5% (C) 6% (D) 7%$
›Reveal solutionSolution
Using $g = 4\pi^2 L / T^2$ and standard error-propagation rules (add relative errors, doubling the exponent's error), the error in $g$ is $1\% + 2\times3\% = 7\%$ .

Concept and Intuition

When a quantity is computed from a formula involving powers of measured quantities, the maximum possible relative (percentage) error in the result is obtained by adding the relative errors of each measured quantity, each multiplied by its power in the formula. This is because errors can compound in the worst case (they don't partially cancel when we want the maximum possible error).

Here, $g = \dfrac{4\pi^2 L}{T^2}$ , so $L$ appears to the power $+1$ and $T$ appears to the power $-2$ (in the denominator, so its magnitude contribution is still $2\times$ the relative error, since $4\pi^2$ is a constant with no error).

Step-by-Step Solution

Write the pendulum formula: $T = 2\pi\sqrt{L/g} \Rightarrow g = \dfrac{4\pi^2 L}{T^2}$ . …
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AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
$Q. The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would: (A) remain unchanged (B) increase towards a saturation value (C) first increase and then decrease to the original value. (D) first decrease and then increase to the original value.$
›Reveal solutionSolution
The pendulum's effective length (distance from pivot to the bob's centre of mass) starts and ends at the same value (the sphere's centre) but dips lower in between as draining water shifts the CM downward, so the period first rises above, then returns to, its original value.

Concept and Intuition

The period of a simple pendulum is $T = 2\pi\sqrt{L/g}$ , where $L$ is the distance from the pivot to the bob's centre of mass. For a uniformly filled sphere (all water, negligible shell mass, or a uniform shell), the CM sits exactly at the sphere's geometric centre — giving the "original" $L$ . As water drains from a hole at the bottom, the remaining water always pools at the bottom of the sphere (gravity), so its own centre of mass is below the sphere's centre, pulling the overall bob CM downward and increasing the effective $L$ (hence increasing $T$ ). Once nearly all the water has drained, what's left is essentially the empty spherical shell again, whose CM (for a thin uniform shell) returns to the geometric centre — the same as the fully-full state — so $T$ returns to its original value.

Step-by-Step Solution

Fully filled: bob CM = sphere's geometric centre $\Rightarrow$ original $L$ , original $T$ .

As water starts leaking, the remaining water occupies only the lower portion of the sphere, so its CM (and hence the whole bob's CM) drops below the centre $\Rightarrow L$ increases $\Rightarrow T$ increases. …
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AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$Q. What is the number of degrees of freedom for an oscillating simple pendulum? (A) One (B) Two (C) Three (D) More than three$
›Reveal solutionSolution
A simple pendulum's bob swings on a fixed-length string in one plane, so its entire state is captured by a single angle $\theta$ — one degree of freedom.

Concept and Intuition

Degrees of freedom = the minimum number of independent coordinates needed to completely specify the configuration (position) of a system. For a simple pendulum, the bob is constrained by an inextensible string of fixed length $L$ to move on a circular arc within a single vertical plane. This constraint reduces what would otherwise be 2 or 3 spatial coordinates down to just one independent variable.

Step-by-Step Solution

A free particle in 3D space has 3 degrees of freedom (x, y, z).

The pendulum bob is constrained to swing in one fixed vertical plane — this removes one degree of freedom, leaving 2 (radial distance from pivot, and angle).

The string is inextensible (fixed length $L$ ), so the radial distance is not free to vary — this removes another degree of freedom. …
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Q.When will the motion of a simple pendulum be simple harmonic?

Why the pendulum is not always simple harmonic

The small-angle approximation

Deriving the period

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