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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 \(\left(\frac{\mathrm{M}}{\mathrm{m}}\right)\) ?

  1. A \(\frac{25}{9}\)
  2. B \(\frac{16}{9}\)
  3. C \(\frac{9}{25}\)
  4. D \(\frac{9}{16}\)
Verified Solution

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 \(M, \frac{25}{9}=1+\frac{m}{M} \quad \therefore \frac{m}{M}=\frac{25}{9}-1=\frac{16}{9}\)
\(\frac{M}{m}=\frac{9}{16}\)