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
Correct — A, √2 T. The period of a simple pendulum is T = 2π√(L/g); it depends only on the length L and the acceleration due to gravity g, and not on the mass of the bob. The second pendulum has the same length but g is halved (g → g/2), so its period is T' = 2π√(L/(g/2)) = 2π√(2L/g) = √2 · T. Doubling the bob's mass changes nothing.
- (b)1/√2 T — This would follow if g were doubled; halving g increases the period, so the factor is √2, not 1/√2.
- (c)2√2 T — No change here gives 2√2; halving g multiplies the period by √2 only, and the mass does not enter the formula.
- (d)T — The period stays the same only if g is unchanged; here g is halved, so the period must change (to √2 T).
A simple pendulum performs simple harmonic motion with time period T = 2π√(L/g). The period is set by the length and the local acceleration due to gravity; for small swings it does not depend on the amplitude, and it never depends on the mass of the bob.
Two changes are offered — mass doubled and g halved. Mass is a decoy because it is absent from the formula. Only the change in g matters, and since T is inversely related to √g, halving g multiplies the period by √2.
- Time period of a simple pendulum: T = 2π√(L/g).
- The period is independent of the bob's mass (and, for small swings, of the amplitude).
- If g is halved while L is fixed, T increases by a factor of √2.
- A pendulum runs slower where g is smaller, such as at greater altitude.
Halving g with L fixed gives T' = √2 T — option (a); the mass is irrelevant.
- Thinking a heavier bob changes the period — mass does not appear in T = 2π√(L/g).
- Getting the direction wrong: smaller g means a longer period, so the factor is √2, not 1/√2.
Asked by changing L, g or the mass and finding the new period as a multiple of the original T.
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. IV. The amplitude of oscillation 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 (SHM). Both questions rest on understanding pendulum oscillation, of which the time-period formula T = 2π√(L/g) is the core result.
The length of a simple pendulum is increased four times to its previous value while the mass is doubled. What is the ratio of the new and previous time period of the pendulum?
- (a) 3 : 1
- (b) 2.5
- (c) 2 : 1
- (d) 3 : 2
Answer(c) 2 : 1
Directly on point — uses T = 2π√(L/g). Quadrupling L multiplies T by √4 = 2, and (as here) doubling the mass has no effect on the period.
- 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 bob
Answer(c) the length and the acceleration due to gravity — T = 2π√(L/g).
- practice — not a real PYQ
If the length of a simple pendulum is made four times, its time period becomes
- (a)half
- (b)double
- (c)four times
- (d)unchanged
Answer(b) double — T is proportional to √L, and √4 = 2.