A steel pendulum clock manufactured at 32^{ci | vedclass

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Question

(A) The time period of a pendulum clock is given by $T = 2\pi \sqrt{\frac{L}{g}}$.The fractional change in time period due to a change in temperature

Vedclass · Accepted Answer

$7.8 \ s$ slow per day

Vedclass · Answer

(A) The time period of a pendulum clock is given by $T = 2\pi \sqrt{\frac{L}{g}}$.The fractional change in time period due to a change in temperature $\Delta 	heta$ is given by $\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta 	heta$, where $\alpha$ is the coefficient of linear expansion.Given: $\alpha = 12 	imes 10^{-6} /{ }^{\circ} C$, $\Delta 	heta = 47^{\circ} C - 32^{\circ} C = 15^{\circ} C$.$\frac{\Delta T}{T} = \frac{1}{2} 	imes 12 	imes 10^{-6} 	imes 15 = 90 	imes 10^{-6}$.The time lost or gained in a day $(86400 \ s)$ is $\Delta t = \frac{\Delta T}{T} 	imes 86400$.$\Delta t = 90 	imes 10^{-6} 	imes 86400 = 7.776 \ s \approx 7.8 \ s$.Since the temperature increases, the length of the pendulum increases, the time period increases, and the clock runs slow.

$A$ steel pendulum clock manufactured at $32^{\circ} C$ and working at $47^{\circ} C$ is nearly (Coefficient of linear expansion of steel $= 12 \times 10^{-6} /{ }^{\circ} C$)

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