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A mass ' $\mathrm{m}_1$ ' is suspended from a spring of negligible mass. A spring is pulled slightly in downward direction and released; mass performs S.H.M. of period ' $\mathrm{T}$ ' '. If the mass is increased by ' $\mathrm{m}_2$ ', the time period becomes ' $\mathrm{T}_2$ '. The ratio $\frac{\mathrm{m}_2}{\mathrm{~m}_1}$ is
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Verified Answer
The correct answer is:
$\frac{\mathrm{T}_2^2-\mathrm{T}_1^2}{\mathrm{~T}_1^2}$
$$
\begin{aligned}
& \mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}_1}{\mathrm{k}}} \text { and } \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{k}}} \\
& \therefore \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\sqrt{\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{~m}_1}} \\
& \therefore \frac{\mathrm{T}_2^2}{\mathrm{~T}_1^2}=\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{~m}_1} \\
& \therefore \frac{\mathrm{T}_2^2-\mathrm{T}_1^2}{\mathrm{~T}_1^2}=\frac{\mathrm{m}_2}{\mathrm{~m}_1}
\end{aligned}
$$
\begin{aligned}
& \mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}_1}{\mathrm{k}}} \text { and } \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{k}}} \\
& \therefore \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\sqrt{\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{~m}_1}} \\
& \therefore \frac{\mathrm{T}_2^2}{\mathrm{~T}_1^2}=\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{~m}_1} \\
& \therefore \frac{\mathrm{T}_2^2-\mathrm{T}_1^2}{\mathrm{~T}_1^2}=\frac{\mathrm{m}_2}{\mathrm{~m}_1}
\end{aligned}
$$
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