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MHT CET · Physics · Oscillations

A body of mass \(64 \mathrm{~g}\) is made to oscillate turn by turn on two different springs A and B. Spring \(\mathrm{A}\) and \(\mathrm{B}\) has force constant \(4 \frac{\mathrm{N}}{\mathrm{m}}\) and \(16 \frac{\mathrm{N}}{\mathrm{m}}\) respectively. If \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) are period of oscillations of springs \(\mathrm{A}\) and \(\mathrm{B}\) respectively then \(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}\) will be

  1. A \(1: 3\)
  2. B \(3: 1\)
  3. C \(1: 2\)
  4. D \(2: 1\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3: 1\)

Step-by-step Solution

Detailed explanation

(A)
\(T=2 \pi \sqrt{\frac{m}{k}} \quad \therefore \frac{T_{1}}{T_{2}}=\sqrt{\frac{k_{2}}{k_{1}}}=\sqrt{\frac{16}{4}}=\frac{2}{1}\)
\(=\frac{T_{1}+T_{2}}{T_{1}-T_{2}}\)