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(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.Taking the logarithmic derivative, we get $\frac{dT}{T} = \frac{1}

Vedclass · Accepted Answer

$2.25 	imes 10^{-6} \, s$

Vedclass · Answer

(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.Taking the logarithmic derivative, we get $\frac{dT}{T} = \frac{1}{2} \frac{dl}{l}$.Since the change in length is given by $\frac{dl}{l} = \alpha \Delta 	heta$, where $\Delta 	heta$ is the change in temperature, we have $\frac{dT}{T} = \frac{1}{2} \alpha \Delta 	heta$.Given $\alpha = 9 	imes 10^{-7} /{ }^{\circ} C$, $\Delta 	heta = 30^{\circ} C - 20^{\circ} C = 10^{\circ} C$, and $T = 0.5 \, s$.The change in time period (loss in time per oscillation) is $dT = T 	imes \frac{1}{2} \alpha \Delta 	heta$.Substituting the values: $dT = 0.5 	imes \frac{1}{2} 	imes 9 	imes 10^{-7} 	imes 10$.$dT = 0.5 	imes 4.5 	imes 10^{-6} = 2.25 	imes 10^{-6} \, s$.

$A$ clock pendulum made of invar has a period of $0.5 \, s$ at $20^{\circ} C$. If the clock is used in a climate where the temperature averages to $30^{\circ} C$, how much time does the clock lose in each oscillation? (For invar, $\alpha = 9 \times 10^{-7} /{ }^{\circ} C$, $g = \text{constant}$)

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