The time period of oscillation of a simple pendulum having length L and mass of the bob m is given as T. If the length of the pendulum is increased to 4L and the mass of the bob is increased to 2m, then which one of the following is the new time period of oscillation?
- (a)T
- (b)2T
- (c)4T
- (d)T/2
Correct — B, 2T. The time period of a simple pendulum is T = 2π√(L/g). The mass of the bob does not appear anywhere in that expression, so doubling it changes nothing — heavier bobs feel a proportionally larger restoring force and have a proportionally larger inertia, and the two effects cancel exactly. Only the length matters, and it enters under a square root, so multiplying L by four multiplies the period by √4, which is 2. The new period is therefore 2T.
- (a)T — This would be right only if neither change mattered. The mass change indeed does not matter, but the length change does — quadrupling the length lengthens the swing measurably, which is why a long clock pendulum ticks more slowly than a short one.
- (c)4T — The trap for anyone who reads the formula without the square root and scales the period in direct proportion to the length. Because L sits inside a square root, a four-fold increase in length gives only a two-fold increase in period.
- (d)T/2 — Right factor, wrong direction. A longer pendulum swings more slowly, not faster, so the period must go up. This option would follow from placing L on top of g inside the root rather than beneath it, or from thinking that the doubled mass speeds the bob up.
For small angular displacements the restoring force on a pendulum bob is proportional to its displacement, which is the condition for simple harmonic motion, and the resulting period is T = 2π√(L/g). Two features of that expression carry most of the exam value. The period is independent of both the mass of the bob and the amplitude of swing — the second of these is Galileo's law of isochronism, and it is why a pendulum could be used as a clock at all. And the period depends on g, so the same pendulum runs slower on a mountain top or on the Moon and faster at sea level.
Two habits solve every question of this family. First, list what the formula does not contain — for a pendulum that is mass and amplitude, so any change to those is a decoy dropped into the stem to see whether you know the formula or only half-remember it. Second, respect the square root: whatever factor multiplies L, the period is multiplied by its square root. Four times the length gives twice the period; nine times gives three times; half the length gives 1/√2. Here the '2m' exists purely to test the first habit and the '4L' purely to test the second.
- The period of a simple pendulum is T = 2π√(L/g) for small oscillations.
- The period is independent of the mass of the bob and, for small swings, of the amplitude.
- Quadrupling the length doubles the period, because length enters under a square root.
- A pendulum about 1 metre long has a period of roughly 2 seconds at the Earth's surface, which is the basis of the seconds pendulum.
- Letting the mass of the bob into the answer. It is absent from the formula entirely.
- Scaling the period in direct proportion to the length and forgetting the square root.
- Assuming a longer pendulum must swing faster because it covers more distance; the period actually lengthens.
NDA asks how the period changes when the length, mass or value of g is altered, or which quantity the period 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
The same oscillator described through its motion rather than through its period — it fixes where the acceleration vanishes and where it peaks, which is the restoring-force picture that the period formula comes from.
A simple pendulum having bob of mass m and length of string l has time period of T. If the mass of the bob is doubled and the length of the string is halved, then the time period of this pendulum will be
- (a) T
- (b) T/√2
- (c) 2T
- (d) √2 T
Answer(b) T/√2
The same question rebuilt with the length halved instead of quadrupled — proof that the mass decoy and the square root are the two things NDA keeps testing here.
If T is the time period of an oscillating pendulum, which one of the following statements is NOT correct ?
- (a) The motion repeats after time T only once
- (b) T is the least time after which motion repeats itself
- (c) The motion repeats itself after nT, where n is a positive integer
- (d) T remains the same only for small angular displacements
Answer(a) The motion repeats after time T only once
From the same year's first session, and it supplies the condition this card's formula quietly assumes — the period is constant only for small angular displacements.
- practice — not a real PYQ
The time period of a simple pendulum is independent of
- (a)its length
- (b)the mass of the bob
- (c)the acceleration due to gravity
- (d)all of these
Answer(b) the mass of the bob — the extra restoring force on a heavier bob is exactly cancelled by its extra inertia.
- practice — not a real PYQ
A simple pendulum of period 2 s is taken to a place where the acceleration due to gravity is one-fourth of its value on the Earth's surface. Its new period is
- (a)1 s
- (b)2 s
- (c)4 s
- (d)8 s
Answer(c) 4 s — g sits in the denominator under the square root, so quartering g doubles the period.