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 answer
Explanation
The correct answer is A, T will increase slightly. Heat makes the copper wire expand a little, and a longer pendulum swings a little more slowly.
The time period of a simple pendulum is T = 2π√(L/g), so it grows with the square root of the length. Copper expands on heating by about 17 parts per million for each degree Celsius. A rise of 30°C therefore lengthens the wire by only about 0.05 per cent, and the period grows by about half of that. This is why old pendulum clocks ran slow in summer and fast in winter, and why makers used invar or compensating pendulums.
B is wrong: the length changes with temperature, so the period cannot stay exactly the same. C is wrong: the wire gets longer, not shorter, on heating. D is wrong: doubling T would need the length to become four times, which a 30°C rise cannot do.
Exam tip: T = 2π√(L/g); heat lengthens the pendulum and the clock loses time.
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