Q.Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is the length of a simple pendulum which ticks seconds ?

Showing the 12 most recent of 20 on this concept.

1. 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.

   …

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2. 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 surfa …

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3. 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.

   …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.Time period of simple pendulum is (A) T = 2π√(l/g) (B) T = √(l/g) (C) T = (1/2π)√(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:

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

   …

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5. 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.

   …

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6. 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:

   …

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7. 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) t …

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8. 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 …

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9. 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 …

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10. 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

    …

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11. 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. …

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12. 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 pendul …

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