Two bodies M and N of equal masses are suspended from two separate massless spring constant k1 and k2 respectively…

Q: Two bodies M and N of equal masses are suspended from two separate massless spring constant k1 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) $\large \frac{k_1}{k_2}$

(b) $\large \sqrt{\frac{k_2}{k_1}}$

(c) $\large \frac{k_2}{k_1}$

(d) $\large \sqrt{\frac{k_1}{k_2}}$

Ans: (b)

Sol: According to question ,

$\large (v_M)_{max} = (v_N)_{max} $

$\large \omega_M A_M = \omega_N A_N $

$\large \frac{A_M}{A_N} = \frac{\omega_N}{\omega_M}$

Since , $\large \omega = \sqrt{\frac{k}{m}} $

$\large \frac{A_M}{A_N} = \sqrt{\frac{k_2}{k_1}}$