Question

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $k_1$ and $k_2$ respectively. lf 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

Answer: (C)

$\sqrt{\frac{k_2}{k_1}}$

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: $v_1=v_2$

$A_1\times \sqrt{\frac{k_1}{m_1}}$ $=A_2\times \sqrt{\frac{k_2}{m_2}}$

$\therefore m_1=m_2⟹\frac{A_1}{A_2}=\sqrt{\frac{k_2}{k_1}}$