MHT CET · Physics · Oscillations
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
- A \(\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2}\)
- B \(\frac{T_1-T_2}{T_1}\)
- C \(\frac{\mathrm{T}_2^2-\mathrm{T}_1^2}{\mathrm{~T}_1^2}\)
- D \(\frac{\mathrm{T}_1^2-\mathrm{T}_2^2}{\mathrm{~T}_1^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{T}_2^2-\mathrm{T}_1^2}{\mathrm{~T}_1^2}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}_1}{\mathrm{k}}} \text { and } \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{k}}} \\
& \therefore \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\sqrt{\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{~m}_1}} \\
& \therefore \frac{\mathrm{T}_2^2}{\mathrm{~T}_1^2}=\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{~m}_1} \\
& \therefore \frac{\mathrm{T}_2^2-\mathrm{T}_1^2}{\mathrm{~T}_1^2}=\frac{\mathrm{m}_2}{\mathrm{~m}_1}
\end{aligned}
\)
\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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