A clock which keeps correct time at 20^circ C | vedclass

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Question

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Taking the derivative,the fractional change in time period is $\fra

Vedclass · Accepted Answer

$10.3$ seconds / day

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Taking the derivative,the fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}$.Since $\Delta l = l \alpha \Delta 	heta$,we have $\frac{\Delta l}{l} = \alpha \Delta 	heta$.Therefore,$\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta 	heta$.Given $\alpha = 12 	imes 10^{-6} /^\circ C$ and $\Delta 	heta = 40^\circ C - 20^\circ C = 20^\circ C$.$\frac{\Delta T}{T} = \frac{1}{2} 	imes 12 	imes 10^{-6} 	imes 20 = 120 	imes 10^{-6} = 1.2 	imes 10^{-4}$.The time lost or gained in a day ($86400$ seconds) is $\Delta T = \frac{\Delta T}{T} 	imes 86400$.$\Delta T = 1.2 	imes 10^{-4} 	imes 86400 = 10.368$ seconds.Since the temperature increases,the length of the pendulum increases,the time period increases,and the clock loses time. Thus,it loses approximately $10.3$ seconds per day.

$A$ clock which keeps correct time at $20^\circ C$ is subjected to $40^\circ C$. If the coefficient of linear expansion of the pendulum is $12 \times 10^{-6} /^\circ C$,how much time will it gain or lose per day?

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