Question

Two springs of spring constants $ {{k}_{1}} $ and $ {{k}_{2}} $ have equal maximum velocities. When executing simple harmonic motion, the ratio of their amplitudes (masses are equal) will be

• A. $ {{\left( \frac{{{k}_{2}}}{{{k}_{1}}} \right)}^{{1}/{2}\;}} $
• B. $ {{\left( \frac{{{k}_{1}}}{{{k}_{2}}} \right)}^{{1}/{2}\;}} $
• C. $ \frac{{{k}_{2}}}{{{k}_{1}}} $
• D. $ {{k}_{1}}{{k}_{2}} $

Answer 1

Answer: (A)

The angular frequency of spring is given by $ \omega =\frac{\sqrt{k}}{m}\propto \sqrt{k} $ (because $ m={{m}_{1}}={{m}_{2}} $ ) For equal maximum velocities, we have $ {{a}_{1}}{{\omega }_{1}}={{a}_{2}}{{\omega }_{2}} $ $ \frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{\omega }_{2}}}{{{\omega }_{1}}}=\frac{\sqrt{{{k}_{2}}}}{{{k}_{1}}}={{\left( \frac{{{k}_{2}}}{{{k}_{1}}} \right)}^{\frac{1}{2}}} $