Q.(a) Derive an expression for time period of oscillation of a Simple Pendulum.
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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.
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=2πgL
That’s it. No mass term. No amplitude term (for small angles).
T=2πgL
Why does this formula make sense?
- L in numerator: longer string → larger T (slower swing). Doubling L multiplies T by 2≈1.4.
- g in denominator: stronger gravity (larger g) → smaller T (faster swing). On the Moon (g≈1.6 m/s²), the same pendulum swings much slower.
- 2π: comes from the mathematics of simple harmonic motion — the pendulum’s motion is approximately sinusoidal for small angles.
To remember: the formula is identical to that of a mass on a spring (T=2πm/k), but here the “restoring force per unit displacement” is mg/L, so the effective “k” is mg/L, giving T=2πL/g.
Common exam pitfalls
- Don’t confuse L with amplitude. L is the string length, not how far you pull it. …
A simple pendulum's period comes from its restoring torque and depends on the local value of gravity, which is why a pendulum clock behaves very differently once that gravity effectively disappears, as inside a freely orbiting spaceship. …
A simple pendulum's period, derived from its restoring torque, is T = 2π√(L/g); since this depends on g, a pendulum clock fails inside an orbiting spaceship, where the effective gravity is zero.
- Consider a simple pendulum of length L and bob of mass m, displaced through a small angle θ from its vertical (equilibrium) position. The restoring force along the arc, tangential to the motion, is due to the component of gravity: F = −mg sinθ For small angles, sinθ ≈ θ (in radians), so: F ≈ −mgθ Since the displacement along the arc is x = Lθ, we have θ = x/L, so: F = −mg(x/L) = −(mg/L) x This is of the form F = −kx (simple harmonic motion), with an effective force constant: k = mg/L For SHM, the angular frequency is ω = √(k/m), so: ω = √[(mg/L)/m] = √(g/L) The time period, T = 2π/ω, is therefore: T = 2π√(L/g)
- No, a pendulum clock cannot show correct time inside a spaceship (in orbit, i.e., in free fall / a state of weightlessness). Inside a freely orbiting or free-falling spaceship, astronauts and all objects (including the pendulum bob) experience zero effective gravity (apparent weightlessness), i.e., effective g ≈ 0. …
Showing the 12 most recent of 20 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.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.
…
- CBSE 2026Set ANNUAL1 markQ.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 …
- CBSE 2026Set ANN1 markQ.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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- CBSE 2025Set ANNUAL1 markMCQQ.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=2πl/g.
For small angular displacement, a simple pendulum of length l undergoes simple harmonic motion under gravity, with restoring torque ≈−mglθ. Solving the resulting SHM equation gives angular frequency ω=g/l, and hence time period:
T=ω2π=2πgl
…
- CBSE 2025Set ANNUAL1 markMCQQ.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=2πl/g, where l is the effective distance from the pivot to the centre of mass of the swinging body.
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- CBSE 2025Set ANNUAL1 markMCQQ.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=2πL/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∝L, a smaller L means a smaller period T.
So:
…
- CBSE 2025Set ANNUAL1 markMCQQ.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 …
- CBSE 2025Set ANNUAL1 markQ.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 …
- CBSE 2024Set ANNUAL1 markMCQQ.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=2πL/g, so T is inversely proportional to 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 …
- CBSE 2024Set ANNUAL1 markQ.What is the length of the second's pendulum?
›Reveal solutionSolution
A seconds pendulum has T=2s; using T=2πL/g gives L≈0.993m, 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=2πgL, so
…
- CBSE 2023Set ANNUAL1 markQ.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=2πL/g contains no amplitude term (for small oscillations), so changing the amplitude does not change T.
For small angular displacements (so that sinθ≈θ), a simple pendulum executes simple harmonic motion with time period
T=2πgL
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 θ0 (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. …
- CBSE 2023Set ANNUAL1 markMCQQ.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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