MHT CET · Physics · Laws of Motion
A light spring is suspended with mass \(m_1\) at its lower end and its upper end fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is \(\mathrm{T}\) second. When a mass \(\mathrm{m}_2\) is added to \(m_1\) and the system is made to oscillate, the period is found to be \(\frac{3}{2} \mathrm{~T}\). The ratio \(\mathrm{m}_1: \mathrm{m}_2\) is
- A \(2:3\)
- B \(3:4\)
- C \(4:5\)
- D \(5:6\)
Answer & Solution
Correct Answer
(C) \(4:5\)
Step-by-step Solution
Detailed explanation
The time period is \(T \propto \sqrt{\mathrm{m}}\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{T}}{3}=\frac{\sqrt{\mathrm{m}_1}}{\sqrt{\mathrm{m}_1+\mathrm{m}_2}} \\
\therefore & \frac{2}{3}=\frac{\sqrt{\mathrm{m}_1}}{\sqrt{\mathrm{m}_1+\mathrm{m}_2}} \\
\therefore \quad & \frac{4}{9}=\frac{\mathrm{m}_1}{\mathrm{~m}_1+\mathrm{m}_2} \\
\therefore \quad & 4 \mathrm{~m}_1+4 \mathrm{~m}_2=9 \mathrm{~m}_1 \\
\therefore \quad & 5 \mathrm{~m}_1=4 \mathrm{~m}_2 \\
\therefore \quad & \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{4}{5}
\end{array}\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{T}}{3}=\frac{\sqrt{\mathrm{m}_1}}{\sqrt{\mathrm{m}_1+\mathrm{m}_2}} \\
\therefore & \frac{2}{3}=\frac{\sqrt{\mathrm{m}_1}}{\sqrt{\mathrm{m}_1+\mathrm{m}_2}} \\
\therefore \quad & \frac{4}{9}=\frac{\mathrm{m}_1}{\mathrm{~m}_1+\mathrm{m}_2} \\
\therefore \quad & 4 \mathrm{~m}_1+4 \mathrm{~m}_2=9 \mathrm{~m}_1 \\
\therefore \quad & 5 \mathrm{~m}_1=4 \mathrm{~m}_2 \\
\therefore \quad & \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{4}{5}
\end{array}\)
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