Question

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $k_1$ and $k_2$ respectively. If the maximum velocities, during oscillations, are equal, the ratio of amplitudes of $A$ and $B$ is:

• (A) $\sqrt{\frac{k_1}{k_2}}$
• (B) $\frac{k_2}{k_1}$
• (C) $\sqrt{\frac{k_2}{k_1}}$
• (D) $\frac{k_1}{k_2}$

Answer 1

The Correct Option is C

Explanation:
Maximum velocity of the particle executing $SHM$ is given by $v=Aw$where $A$ is the amplitude and $w=\sqrt{\frac{k}{m}}$Let the amplitude of the particle $A$ and $B$ be $A_1$ and $A_2$ respectively.Given: $v_1=v_2$$A_1\times \sqrt{\frac{k_1}{m_1}}=A_2\times \sqrt{\frac{k_2}{m_2}}$$\therefore m_1=m_2$$⟹\frac{A_1}{A_2}=\sqrt{\frac{k_2}{k_1}}$Hence, the correct option is (C).