The time period of a 1 m long pendulum approximates to
- (a)6 s
- (b)4 s
- (c)2 s
- (d)1 s
Correct — C, 2 s. A simple pendulum's time period is T = 2π√(L/g). For L = 1 m and g ≈ 9·8 m/s², T = 2π√(1/9·8) ≈ 2π × 0·319 ≈ 2·0 s.
- (a)6 s — 6 s is roughly three times the correct value; since T ∝ √L, that would need a pendulum about 9 m long, not 1 m.
- (b)4 s — 4 s would require a length of about 4 m (because T ∝ √L), not 1 m.
- (d)1 s — A period of 1 s corresponds to a length of about 0·25 m; a 1 m pendulum gives roughly 2 s, so 1 s is too short.
For small swings, a simple pendulum's time period depends only on its length L and the local gravity g: T = 2π√(L/g). It is independent of the bob's mass and, for small angles, of the amplitude. A 1 m pendulum has T ≈ 2 s — close to the 'seconds pendulum'.
Plug L = 1 and g ≈ 9·8 (or 10) into T = 2π√(L/g). Because T grows only as √L, quadrupling the length merely doubles the period — which is why 4 s and 6 s, needing much longer pendulums, are wrong.
- Time period of a simple pendulum: T = 2π√(L/g).
- A 1 m long pendulum has a period of approximately 2 s.
- The period is independent of the bob's mass and, for small angles, of the amplitude.
- A pendulum whose period is 2 s (one second per swing) is called a 'seconds pendulum', about 0·99 m long.
- Thinking the period depends on the bob's mass — it does not.
- Forgetting the square-root: the period grows as √L, not as L.
A direct calculation of T from L and g, or a concept check on what the period does and does not depend on.
Consider the following statements: A simple pendulum is set into oscillation. Then I. The acceleration is zero when the bob passes through the mean position. II. In each cycle the bob attains a given velocity twice. III. Both acceleration and velocity of the bob are zero when it reaches its extreme position during its oscillation. IV. The amplitude of oscillation of the simple pendulum decreases with time. Which of these statements are correct?
- (a) I and II
- (b) III and IV
- (c) I, II and IV
- (d) II, III and IV
Answer(c) I, II and IV
Same concept — the motion of a simple pendulum. This 2022 item uses the period formula T = 2π√(L/g); the 2001 UPSC item probes the related simple-harmonic behaviour of the same pendulum (acceleration at the mean position, damping of amplitude).
A pendulum clock is lifted to a height where the gravitational acceleration has a certain value g. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is g/2. The time period of the second pendulum would be (in terms of period T of the first pendulum)
- (a) √2 T
- (b) 1/√2 T
- (c) 2√2 T
- (d) T
Answer(a) √2 T
Same formula — T = 2π√(L/g). This 2022 item computes the period for a 1 m pendulum; the 2019 item tests how that same period depends on g (and that the bob's mass has no effect).
Which one of the following statements regarding simple pendulum is correct? Simple pendulum has a time period independent of amplitude:
- (a) only for small amplitudes because then the net force on its bob is independent of its displacement.
- (b) for any amplitude because the net force on the bob is always proportional to its displacement.
- (c) for any amplitude because the net force on the bob is independent of its displacement.
- (d) only for small amplitudes because then the net force on its bob is proportional to its displacement.
Answer(d) only for small amplitudes because then the net force on its bob is proportional to its displacement.
Related — the conditions behind the period formula. It explains why T = 2π√(L/g) (used in this 2022 item) holds only for small-angle swings, where the motion is simple harmonic.
- practice — not a real PYQ
To double the time period of a simple pendulum, its length must be made
- (a)twice
- (b)four times
- (c)half
- (d)eight times
Answer(b) four times — T ∝ √L, so the length must be 4× for the period to double.
- practice — not a real PYQ
The time period of a simple pendulum depends on
- (a)the mass of the bob
- (b)the amplitude for large swings
- (c)the length and the acceleration due to gravity
- (d)the material of the string
Answer(c) the length and the acceleration due to gravity — T = 2π√(L/g).