The time period of a simple pendulum made using a thin copper wire of length L is T. Suppose the temperature of the room in which this simple pendulum is placed increases by 30°C, what will be the effect on the time period of the pendulum ?
- (a)T will increase slightly
- (b)T will remain the same
- (c)T will decrease slightly
- (d)T will become more than 2 times
Correct — A, T will increase slightly. The time period of a simple pendulum is T = 2π√(L/g), so it depends on the length of the suspension and on the local value of g, and on nothing else — not on the mass of the bob and not, for small swings, on the amplitude. Warming the room makes the copper wire longer, because a solid expands on heating; L grows, and since T varies as the square root of L, T grows too. The growth is small: copper expands by roughly 17 parts in a million for each degree Celsius, so 30 °C lengthens the wire by about five parts in ten thousand, and the period rises by about half of that fraction. Small, but in the direction of an increase — which is why a pendulum clock loses time in hot weather.
- (b)T will remain the same — This would be right only if the wire kept its length. Metals expand measurably on heating, and the length of the suspension is the one dimension the period cares about, so the period cannot be entirely unaffected.
- (c)T will decrease slightly — A decrease would require the wire to shorten, which is what cooling does. The question raises the temperature, so the length rises and the period rises with it.
- (d)T will become more than 2 times — Because T varies as the square root of L, doubling the period would need the length to become four times what it was. No metal expands by anything close to that over a 30 °C rise — the change here is a fraction of a thousandth.
A simple pendulum swinging through a small angle performs simple harmonic motion with period T = 2π√(L/g). Two things can therefore change its period — the length of the suspension and the acceleration due to gravity where it hangs. Thermal expansion acts on the first of these: heating lengthens the metal suspension and cools shortens it, so a pendulum clock runs slow in summer and fast in winter unless it is compensated.
The question is really two ideas joined together. First, know what the period depends on and what it does not; the mass of the bob is a favourite red herring and is absent from the formula. Second, know the direction of thermal expansion and the rough size of it. Once you see that L rises by a tiny fraction, the square root guarantees that T rises by an even tinier one, which rules out the dramatic option and leaves only the 'slightly' answer in the right direction. Clockmakers solved the same problem long ago with the gridiron pendulum and later with invar rods, whose expansion is very small.
- For small oscillations the period of a simple pendulum is T = 2π√(L/g), independent of the mass of the bob.
- Solids expand on heating; the linear expansion coefficient of copper is roughly 17 × 10⁻⁶ per degree Celsius.
- Because T varies as the square root of L, a fractional rise in length produces about half that fractional rise in period.
- Pendulum clocks are compensated by gridiron pendulums or low-expansion alloys such as invar so that summer heat does not make them lose time.
- Bringing the mass of the bob into the period — it does not appear in the formula.
- Reading 'increases by 30 °C' as a large change; for a metal it is a change of a few parts in ten thousand.
- Forgetting the square root, and so expecting the period to change in the same proportion as the length.
NDA asks how the period responds when length, mass, amplitude or g is altered, and sometimes wraps the change inside a temperature rise as here.
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 formula tested by changing the two things a candidate is most likely to confuse, with the mass planted as the decoy exactly as it is here.
- practice — not a real PYQ
A pendulum clock keeping correct time in winter is taken into a hot room. It will
- (a)gain time because the rod contracts
- (b)lose time because the rod lengthens
- (c)keep correct time because the bob is unchanged
- (d)stop swinging altogether
Answer(b) lose time because the rod lengthens — a longer pendulum has a longer period, so it completes fewer swings in a day.
- practice — not a real PYQ
The length of a simple pendulum is made four times its original value. Its time period becomes
- (a)half the original
- (b)the same as the original
- (c)twice the original
- (d)four times the original
Answer(c) twice the original — the period varies as the square root of the length, and the square root of four is two.