Question

Two bodies M and N of equal masses are suspended from two separate massless springs of spring constants kx and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the one amplitude of vibration of M to that of N is

• A. $k_1/k_2$
• B. $\sqrt{k_2/k_1}$
• C. $k_2/k_1$
• D. $\sqrt{k_1/k_2}$

Answer 1

Answer: (B)

$\sqrt{k_2/k_1}$

Answer 2

$(v_M)_{max} = (v_N)_{max}$ $\therefore \, \, \, \, {\omega}_M A_M = {\omega}_N A_N$ or $\, \, \, \, \, \, \, \, \, \, \, \, \frac{A_M}{ A_N} = \frac{{\omega}_M}{{\omega}_N} = \sqrt{\frac{k_2}{k_1}} \, \, \, \, \, \bigg( \because \omega = \sqrt{\frac{k}{m}} \bigg)$ $\therefore \, \,$ Correct answer is (b).