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 oscillation, are equal, the ratio of amplitude 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)

Maximum velocity during SHM = A$\omega$
But k = m$\omega^{2}$
$\therefore \,\omega = \sqrt{\frac{k}{m}}$
$\therefore \,$ Maximum velocity $= A\sqrt{\frac{k}{m}}$
Here the maximum velocity is same and m is also same
$\therefore \, A_{1}\sqrt{k_{1}}=A_{2}\sqrt{k_{2}}\quad\quad\quad\therefore \, \frac{A_{1}}{A_{2}}=\sqrt{\frac{k_{2}}{k_{1}}}$