MHT CET · Physics · Oscillations
A mass 'M' is suspended from a spring of negligible mass. The spring is pulled a little
and then released so that the mass executes S.H.M. of period \(\mathrm{T}\). If the mass is increased by 'm', the time period becomes \(\frac{5 \mathrm{~T}^{\prime}}{3}\). What is the ratio \(\left(\frac{\mathrm{M}}{\mathrm{m}}\right)\) ?
- A \(\frac{25}{9}\)
- B \(\frac{16}{9}\)
- C \(\frac{9}{25}\)
- D \(\frac{9}{16}\)
Answer & Solution
Correct Answer
(D) \(\frac{9}{16}\)
Step-by-step Solution
Detailed explanation
\(T=2 \pi \sqrt{\frac{M}{k}} ; \quad \frac{5 T}{3}=2 \pi \sqrt{\frac{M+m}{k}}\)
\(\therefore \frac{5}{3} \times 2 \pi \sqrt{\frac{M}{k}}=2 \pi \sqrt{\frac{M+m}{k}}\)
\(\therefore \frac{25}{9} \cdot \frac{M}{k}=\frac{M+m}{k}\)
\(\therefore \frac{25}{9} M=M+m\)
Dividing by \(M, \frac{25}{9}=1+\frac{m}{M} \quad \therefore \frac{m}{M}=\frac{25}{9}-1=\frac{16}{9}\)
\(\frac{M}{m}=\frac{9}{16}\)
\(\therefore \frac{5}{3} \times 2 \pi \sqrt{\frac{M}{k}}=2 \pi \sqrt{\frac{M+m}{k}}\)
\(\therefore \frac{25}{9} \cdot \frac{M}{k}=\frac{M+m}{k}\)
\(\therefore \frac{25}{9} M=M+m\)
Dividing by \(M, \frac{25}{9}=1+\frac{m}{M} \quad \therefore \frac{m}{M}=\frac{25}{9}-1=\frac{16}{9}\)
\(\frac{M}{m}=\frac{9}{16}\)
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