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.
Correct — D, the period is amplitude-independent only for small amplitudes, because only then is the net restoring force proportional to the bob's displacement. The force pulling the bob back toward the mean position is the tangential component of gravity, F = -mg sin(theta), where theta is the angular displacement. When theta is small, sin(theta) is approximately theta, so the force becomes proportional to the displacement (F = -(mg/L)x for arc displacement x = L*theta). A force proportional to displacement is the defining condition of simple harmonic motion, whose period T = 2*pi*sqrt(L/g) depends only on length and gravity, not on how far the bob swings. At large amplitudes sin(theta) can no longer be replaced by theta, the force is no longer proportional to displacement, and the period grows with amplitude.
- (a)only for small amplitudes because then the net force on its bob is independent of its displacement. — The amplitude condition is right, but the reason is inverted — at small amplitudes the restoring force is proportional to the displacement, not independent of it. A force that did not vary with displacement could not produce oscillation at all.
- (b)for any amplitude because the net force on the bob is always proportional to its displacement. — The restoring force -mg sin(theta) is proportional to displacement only in the small-angle limit; for large swings sin(theta) is not equal to theta, so the force is not proportional to displacement and the period does depend on amplitude.
- (c)for any amplitude because the net force on the bob is independent of its displacement. — Both halves fail — the restoring force clearly depends on displacement (that is what makes it a restoring force), and the period is not amplitude-independent for large swings.
A simple pendulum behaves as a simple harmonic oscillator only approximately. Its motion is simple harmonic — and its period fixed regardless of amplitude — solely in the small-angle regime, where the sine of the angular displacement can be replaced by the angle itself.
The tempting error is Galileo's idealised isochronism — the belief that a pendulum's period is the same for every swing. That holds only for small oscillations; the question rewards knowing both why (force proportional to displacement) and when (small amplitudes).
- The restoring force on the bob is the tangential component of gravity, F = -mg sin(theta), directed toward the mean position.
- For small angles sin(theta) is approximately theta (in radians), so the force becomes proportional to displacement — the condition for simple harmonic motion.
- In that regime the period is T = 2*pi*sqrt(L/g), set only by the pendulum's length and the acceleration due to gravity, independent of amplitude and of the bob's mass.
- The small-angle approximation is customarily treated as valid up to roughly 10 to 15 degrees; beyond that the period lengthens noticeably with amplitude.
The period is independent of amplitude only when the restoring force stays proportional to displacement — the small-angle case.
- Assuming a pendulum's period is amplitude-independent for every swing — that isochronism holds only in the small-angle limit.
- Thinking the period depends on the bob's mass — it does not; it depends on length and g.
Framed as a which-statement-is-correct item that pairs the amplitude condition with its cause — match 'small amplitudes' with 'force proportional to displacement'.
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 — acceleration is zero at the mean position, a given speed is reached twice each cycle, and a real pendulum's amplitude decays with time; only statement III (both zero at the extreme) is false.
Both probe the mechanics of a simple pendulum's oscillation — the UPSC item tests where acceleration and velocity vanish and that real pendulums lose amplitude over time, the same SHM framework in which this NDA question about amplitude-independence sits.
- practice — not a real PYQ
For small oscillations, the time period of a simple pendulum of length L is proportional to:
- (a)L
- (b)sqrt(L)
- (c)L squared
- (d)1/sqrt(L)
Answer(b) sqrt(L) — since T = 2*pi*sqrt(L/g), the period varies as the square root of the length.
- practice — not a real PYQ
A simple pendulum is taken from the Earth to the surface of the Moon, where g is about one-sixth of its value on Earth. Its time period will:
- (a)decrease
- (b)remain the same
- (c)increase
- (d)become zero
Answer(c) increase — T = 2*pi*sqrt(L/g), so a smaller g makes the period longer (about sqrt(6) times).