The length of a second's pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on surface of planet X such that the acceleration of the planet X is $n$ times greater than the Earth is a) $0.9n$ m b) $0.9/n$ m c) $0.9n^2$ m d) $0.9/n^2$ m

The length of a second's pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on sur

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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.
Don’t include mass. The period does not depend on the bob’s mass — a common trick question.
Small-angle assumption. If the problem says “small oscillations” or gives an angle < 10°, use this formula. If the angle is large, the formula changes (and is rarely asked in basic exams).

Quick check

A pendulum of length 1 m on Earth ( $g = 9.8$ m/s²) has period:

T = 2\pi \sqrt{\frac{1}{9.8}} \approx 2\pi \times 0.319 \approx 2.0 \text{ seconds}

That’s why a “seconds pendulum” (period exactly 2 s) has length about 1 m — a classic exam fact.

Final answer: For a simple pendulum with small amplitude, the period is $T = 2\pi \sqrt{L/g}$ , independent of mass and amplitude.

Many students find this page while searching "Simple Pendulum Period formula physics" or "Simple Pendulum Period important questions and answers"; the concept sits firmly within the Class 11 Physics NCERT/CBSE syllabus. It's also a frequent building block for numericals in JEE Main, NEET and state engineering/medical entrance exams, so treating it as a one-time memorisation task rather than an understood idea tends to backfire later.

Step 1. A 'seconds pendulum' is, by definition, one whose time period is exactly $T=2$ s, on Earth or on planet X alike, since $T$ is fixed at 2 s in both cases.

Step 2. From $T=2\pi\sqrt{l/g}$ , squaring gives $l/g=(T/2\pi)^2$ , a constant, the same on Earth and on planet X since $T=2$ s on both.

Step 3. So $\dfrac{l_{\text{Earth}}}{g_{\text{Earth}}}=\dfrac{l_X}{g_X}$ , giving $l_X=l_{\text{Earth}}\times\dfrac{g_X}{g_{\text{Earth}}}$ .

Step 4. Given $g_X=n\,g_{\text{Earth}}$ and $l_{\text{Earth}}=0.9$ m, $l_X=0.9\times n=0.9n$ m.

✓Final answer

0.9n

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 20 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. A child swinging on a Swing in the sitting position stands up, then the time period of the Swing will :$
(a) Increase
(b) Decrease
(c) Remain the same
(d) Increase, if the Child is long and decrease, if the Child is short.
›Reveal solutionSolution

Standing up raises the child's centre of mass, shortening the effective pendulum length, so by T = 2π√(l/g) the time period decreases.

A swing with a child on it behaves approximately like a simple pendulum, with time period:

T = 2π√(l/g)

where l is the effective length — the distance from the point of support (pivot) to the centre of mass of the child + swing system.

When the child is sitting, the centre of mass is lower (farther from the pivot), so l is larger.

When the child stands up, their centre of mass shifts upward (closer to the pivot), decreasing the effective length l.

Since T is proportional to √l, a smaller l means a smaller T — the time period decreases.

✓Final answer
$The correct option is (b) Decrease.$
CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: The time period of a second pendulum remains ____ second.$
›Reveal solutionSolution

A second's pendulum has a time period of exactly 2 seconds by definition.

A second's pendulum is a simple pendulum whose length is specifically tuned so that it takes exactly 1 second to swing from one extreme position to the other (a half-oscillation), meaning its full time period (one complete to-and-fro oscillation) is 2 seconds. Using T = 2π sqrt(l/g), at Earth's surface (g ≈ 9.8 m/s²) this corresponds to a pendulum length of approximately 0.994 m (about 1 m).

✓Final answer
$The time period of a second's pendulum remains 2 seconds.$
CBSE 2026Set ANN1 mark
$Q. The simple pendulum whose time period of oscillation is 2 seconds is called a ______ .$
›Reveal solutionSolution

A simple pendulum with a time period of 2 seconds is called a seconds pendulum.

A seconds pendulum is defined as a simple pendulum whose time period of oscillation is exactly 2 seconds. This means it takes 1 second to swing from one extreme to the other, so it 'ticks' once every second - which is why it is used in pendulum clocks.

Its length can be found from T = 2 pi sqrt(l/g): with T = 2 s and g = 9.8 m/s^2, l is about 0.99 m (close to 1 metre).

✓Final answer
$Seconds pendulum.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Time period of simple pendulum is (A) T = 2π\sqrt(l/g) (B) T = \sqrt(l/g) (C) T = (1/2π)\sqrt(l/g) (D) T = g/(2πl)$
›Reveal solutionSolution

The time period of a simple pendulum is $T = 2\pi\sqrt{l/g}$ .

For small angular displacement, a simple pendulum of length $l$ undergoes simple harmonic motion under gravity, with restoring torque $\approx -mgl\theta$ . Solving the resulting SHM equation gives angular frequency $\omega = \sqrt{g/l}$ , and hence time period:

$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{l}{g}}$

This shows $T$ depends only on the pendulum's length $l$ and local $g$ — not on the mass of the bob or the amplitude (for small oscillations).

✓Final answer
$(A) T = 2π\sqrt(l/g).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. A girl is sitting on a swing and swinging. If she stands up then time period of the swing will (A) increase (B) decrease (C) remain same (D) none of these$
›Reveal solutionSolution

When the girl stands up, the swing's time period decreases.

A swing behaves approximately as a simple pendulum, with time period $T = 2\pi\sqrt{l/g}$ , where $l$ is the effective distance from the pivot to the centre of mass of the swinging body.

When the girl stands up on the swing, her centre of mass shifts upward (closer to the pivot/support), which reduces the effective length $l$ of the pendulum. Since $T \propto \sqrt{l}$ , a smaller $l$ gives a smaller time period.

✓Final answer
$(B) decrease.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Assertion (A): When a girl sitting on a swing stands up, the periodic time of the swing will increase. Reason (R): In standing position of the girl, the length of swing will increase. Select the correct answer from the codes below.$
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) and (R) both are false.
›Reveal solutionSolution
Standing up moves the person's centre of mass closer to the pivot, shortening (not lengthening) the effective pendulum length — so the period decreases (this is exactly how children 'pump' a swing to go faster).

A swing with a person on it behaves like a simple pendulum, with period:

$T = 2\pi\sqrt{L/g}$

where $L$ is the distance from the pivot to the person's centre of mass.

When the girl stands up on the swing, her centre of mass moves upward, i.e. closer to the pivot point — this decreases the effective length $L$ , not increases it. Since $T \propto \sqrt{L}$ , a smaller $L$ means a smaller period $T$ .

So:

Assertion (A) claims the period increases — this is false (it decreases).

Reason (R) claims the effective length increases — this is also false (it decreases).

This is the same physical principle behind how a person standing/crouching rhythmically on a swing pumps energy into it to swing higher and faster.

✓Final answer
$Both (A) and (R) are false — option (d) .$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Match the following - Column A item: Time period of simple pendulum. Pick the matching relation from Column B.$
(a) F.V (Force . Velocity)
(b) T is proportional to sqrt(l)
(c) eta is proportional to 1/(dv/dx)
(d) mu_s = tan(theta)
(e) Y is proportional to 1/l
(f) v^2/r
›Reveal solutionSolution

The time period of a simple pendulum, T = 2pisqrt(l/g), is proportional to the square root of its length l.

For small oscillations, a simple pendulum of length l undergoes SHM with time period T = 2pisqrt(l/g), where g is the acceleration due to gravity. Since g is constant at a given location, T is directly proportional to sqrt(l) — a longer pendulum takes proportionally longer (by the square root) to complete one oscillation, matching the printed relation T proportional to sqrt(l).

✓Final answer
$Time period of simple pendulum matches option (b): T is proportional to sqrt(l).$
CBSE 2025Set ANNUAL1 mark
$Q. State True or False: The time period of second pendulum is 1 second.$
›Reveal solutionSolution

The statement is False: a seconds pendulum has a time period of 2 seconds.

A 'second's pendulum' is, by definition, a simple pendulum whose time period is exactly 2 seconds — it completes one full oscillation (there and back) in 2 seconds. It appears to 'tick' once every second because each tick corresponds to the pendulum passing through its mean position, which happens twice per full oscillation (once moving each way), i.e., once every half-period = once every 1 second. So while it ticks every 1 second, its actual time period (duration of one complete oscillation) is 2 seconds, not 1 second.

✓Final answer
$False — the time period of a second's pendulum is 2 seconds.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The time-period of a simple pendulum (A) is infinite at poles (B) is more at poles than equator (C) is same at both places (D) is more at equator than poles$
›Reveal solutionSolution

Since $g$ is smaller at the equator, a pendulum's period is longer there than at the poles.

Time period: $T=2\pi\sqrt{L/g}$ , so $T$ is inversely proportional to $\sqrt g$ . Due to earth's equatorial bulge and rotation, the value of $g$ is smaller at the equator than at the poles. A smaller $g$ gives a larger $T$ , so the pendulum's period is greater at the equator than at the poles.

✓Final answer
$(D) is more at equator than poles.$
CBSE 2024Set ANNUAL1 mark
$Q. What is the length of the second's pendulum?$
›Reveal solutionSolution

A seconds pendulum has $T=2\,s$ ; using $T=2\pi\sqrt{L/g}$ gives $L \approx 0.993\,m$ , i.e. about 1 metre.

A second's pendulum is defined as a simple pendulum whose time period is exactly 2 seconds (so it takes 1 second to swing from one extreme to the other). The time period of a simple pendulum is $T = 2\pi\sqrt{\dfrac{L}{g}}$ , so

$L = \dfrac{gT^2}{4\pi^2} = \dfrac{9.8 \times (2)^2}{4\pi^2} = \dfrac{39.2}{39.48} \approx 0.993\,m$

✓Final answer
$0.993\,m$
CBSE 2023Set ANNUAL1 mark
$Q. What will be the value of the time period of a simple pendulum if the amplitude is reduced to half of its original value?$
›Reveal solutionSolution

A simple pendulum's time period $T = 2\pi\sqrt{L/g}$ contains no amplitude term (for small oscillations), so changing the amplitude does not change $T$ .

For small angular displacements (so that $\sin\theta \approx \theta$ ), a simple pendulum executes simple harmonic motion with time period

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

where $L$ is the effective length of the pendulum and $g$ is the acceleration due to gravity. This expression depends only on $L$ and $g$ — it does not contain the amplitude $\theta_0$ (or the amplitude of linear displacement) at all. This is precisely the defining property of simple harmonic motion: isochronism, i.e. the period is independent of amplitude as long as the oscillations are small. So if the amplitude is reduced to half its original value (while remaining small), $L$ and $g$ are unaffected, and the time period stays exactly the same as before.

✓Final answer
$T = 2\pi\sqrt{L/g}$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Time period of the second's pendulum is:$
(a) 1 second
(b) 2 second
(c) 3 second
(d) Infinite
›Reveal solutionSolution

By definition, a 'second's pendulum' is a simple pendulum with a time period of exactly 2 seconds.

A seconds pendulum is designed so that it ticks (completes one swing from one extreme to the other) every second -- meaning a full oscillation (there and back) takes 2 seconds. Using T = 2pisqrt(l/g), setting T = 2 s and g = 9.8 m/s^2 gives the length of such a pendulum as approximately 0.994 m, close to 1 metre -- this is why the second's pendulum is a classic length standard.

✓Final answer
$The correct option is (b) 2 second.$

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Q.The length of a second's pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on surface of planet X such that the acceleration of the planet X is nnn times greater than the Earth is a) 0.9n0.9n0.9n m b) 0.9/n0.9/n0.9/n m c) 0.9n20.9n^20.9n2 m d) 0.9/n20.9/n^20.9/n2 m

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Q.The length of a second's pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on surface of planet X such that the acceleration of the planet X is $n$ times greater than the Earth is a) $0.9n$ m b) $0.9/n$ m c) $0.9n^2$ m d) $0.9/n^2$ m

Unlock everything free for 14 days