Question

Two bodies M and N of equal masses are suspended from two separate massless springs of spring constants $ {{k}_{1}} $ and $ {{k}_{2}} $ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitudes of vibrations of M to that of N is

• A. $ \frac{{{k}_{1}}}{{{k}_{2}}} $
• B. $ \sqrt{\frac{{{k}_{1}}}{{{k}_{2}}}} $
• C. $ \frac{{{k}_{2}}}{{{k}_{1}}} $
• D. $ \sqrt{\frac{{{k}_{2}}}{{{k}_{1}}}} $

Answer 1

Answer: (D)

Maximum velocities are equal. Hence, $ {{a}_{1}}{{\omega }_{1}}={{a}_{2}}{{\omega }_{2}} $ $ \frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{\omega }_{2}}}{{{\omega }_{1}}} $ $ =\frac{2\pi /{{n}_{2}}}{2\pi /{{n}_{1}}} $ $ =\frac{{{n}_{1}}}{{{n}_{2}}} $ $ =\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{2\pi \sqrt{\frac{m}{{{k}_{1}}}}}{2\pi \sqrt{\frac{m}{{{k}_{2}}}}}=\sqrt{\frac{{{k}_{2}}}{{{k}_{1}}}} $