Question

Two bodies $A$ and $B$ of equal masses are suspended from two separate springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate such that their maximum velocities are equal, the ratio of the amplitudes of oscillation of $A$ and $B$ will be

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

Answer 1

The correct option is D $\sqrt{\frac{k_2}{k_1}}$

The velocity of an oscillating body is maximum when it is at the equilibrium position where the potential energy is zero and the energy is entirely kinetic. At the extreme positions, the kinetic energy is zero and the energy is entirely potential.

Therefore, the kinetic energy at equilibrium position = potential energy at extreme positions = total energy.

Since the maximum velocities (i.e. velocities at equilibrium position) are equal for the two equal masses, their kinetic energies are also equal. Which means their potential energies at extreme positions where the displacement is maximum are equal.
If $x_1$ and $x_2$ are the amplitudes of bodies $A$ and $B$, we have
$\frac{1}{2}{k}_1{x}_1^2=\frac{1}{2}{k}_2{x}_2^2$
or $\frac{x_1}{x_2}=\sqrt{\frac{k_2}{k_1}}$
Hence, the correct choice is (d).