Question

An insect of negligible mass is sitting on a block of mass $M$, tied with a spring of force constant $k$. The block performs simple harmonic motion with amplitude $A$ in front of a plane mirror as shown. The maximum speed of the insect relative to its image will be

• A. $A\sqrt{\frac{k}{M}}$
• B. $\frac{A\sqrt{3}}{2}\sqrt{\frac{k}{M}}$
• C. $A\sqrt{3}\sqrt{\frac{k}{M}}$
• D. $2A\sqrt{\frac{M}{k}}$

Answer 1

The correct option is C $A\sqrt{3}\sqrt{\frac{k}{M}}$

As the insect sits on the block, the maximum speed of the insect will be equal to the maximum speed of the block.

$v=\omega A$

$\Rightarrow v=\sqrt{\frac{k}{M}}A........\left(1\right)$

Assuming the block is at equilibrium position and going downward, the components of its speed along and perpendicular to the plane mirror are as shown in the given figure.


From the figure, velocity of object (insect)
$\overrightarrow{v_o}=v\cos {60}^{\circ }\hat{i}+v\sin {60}^{\circ }\hat{j}$

[assuming the axes as shown]

We know that, in case of stationary plane mirror, velocity of image is same as velocity of object along the plane mirror, but for the direction perpendicular to the plane mirror, velocity of image is in opposite direction to that of the object, but with same magnitude.

Thus, velocity of image,
$\overrightarrow{v_i}=v\cos {60}^{\circ }\hat{i}-v\sin {60}^{\circ }\hat{j}$

Velocity of insect w.r.t its image
$\overrightarrow{v_{o,i}}=\overrightarrow{v_o}-\overrightarrow{v_i}=v\cos {60}^{\circ }\hat{i}+v\sin {60}^{\circ }\hat{j}-(v\cos {60}^{\circ }\hat{i}-v\sin {60}^{\circ }\hat{j})$

$\Rightarrow \overrightarrow{v_{o,i}}=2v\sin {60}^{\circ }\hat{j}$

Magnitude of relative velocity, $|\overrightarrow{v_{o,i}}|=2v\sin {60}^{\circ }$
$=2A\sqrt{\frac{k}{M}}\times \frac{\sqrt{3}}{2}$ [from Eq.(1)]
$\Rightarrow |\overrightarrow{v_{o,i}}|=\sqrt{3}A\sqrt{\frac{k}{M}}$

Hence, option $\left(c\right)$ is the correct answer.