Question
A torsional pendulum consists of a solid disc connected to a thin wire$\Big(\alpha=2.4\times10^{-5}\ ^\circ\text{C}^{-1}\Big)$at its centre. Find the percentage change in the time period between peak winter (5°C) and peak summer (45°C).
✓
Answer
Let the initial m.I. at 0°C be $\text{I}_0$
$\text{T}=2\pi\sqrt\frac{\text{I}}{\text{K}}$
$\text{I}=\text{I}_0(1+2\alpha\Delta\theta)$ (from above question)
At 5°C, $\text{T}_1=2\pi\sqrt\frac{\text{I}_0(1+2\alpha\Delta\theta)}{\text{K}}$
$=2\pi\sqrt\frac{\text{I}_0(1+2\alpha\Delta5)}{\text{k}}$
$=2\pi\sqrt\frac{\text{I}_0(1+10\alpha)}{\text{K}}$
At 45°C, $\text{T}_2=2\pi\sqrt\frac{\text{I}_0(1+2\alpha45)}{\text{K}}$
$2\pi\sqrt\frac{\text{I}_0(1+90\alpha)}{\text{K}}$
$\frac{\text{T}_2}{\text{T}_1}=\sqrt\frac{1+90\alpha}{1+10\alpha}$
$=\sqrt\frac{1+90\times2.4\times10^{-5}}{1+10\times2.4\times10^{-5}}\sqrt\frac{1.00216}{1.00024}$
% change $=\Big(\frac{\text{T}_2}{\text{T}_1}-1\Big)\times100$
$=0.0959\%=9.6\times10^{-2}\%$
$\text{T}=2\pi\sqrt\frac{\text{I}}{\text{K}}$
$\text{I}=\text{I}_0(1+2\alpha\Delta\theta)$ (from above question)
At 5°C, $\text{T}_1=2\pi\sqrt\frac{\text{I}_0(1+2\alpha\Delta\theta)}{\text{K}}$
$=2\pi\sqrt\frac{\text{I}_0(1+2\alpha\Delta5)}{\text{k}}$
$=2\pi\sqrt\frac{\text{I}_0(1+10\alpha)}{\text{K}}$
At 45°C, $\text{T}_2=2\pi\sqrt\frac{\text{I}_0(1+2\alpha45)}{\text{K}}$
$2\pi\sqrt\frac{\text{I}_0(1+90\alpha)}{\text{K}}$
$\frac{\text{T}_2}{\text{T}_1}=\sqrt\frac{1+90\alpha}{1+10\alpha}$
$=\sqrt\frac{1+90\times2.4\times10^{-5}}{1+10\times2.4\times10^{-5}}\sqrt\frac{1.00216}{1.00024}$
% change $=\Big(\frac{\text{T}_2}{\text{T}_1}-1\Big)\times100$
$=0.0959\%=9.6\times10^{-2}\%$
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