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 infront of a plane mirror placed as shown. The maximum speed of 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. $A\sqrt{\frac{M}{K}}$

Answer 1

Answer: (C)

$A\sqrt{3}\,\sqrt{\frac{K}{M}}$

Answer 2

The insect moves in SHM with the block, so its maximum speed is $v_{\text{max}} = A\sqrt{\frac{K}{M}}$.

In a plane mirror, the image’s displacement normal to the mirror is the same as that of the object but in the opposite direction. Hence, the image’s velocity component normal to the mirror is equal in magnitude to the insect’s.

The relative speed between the insect and its image is twice the component of the insect's speed along the mirror normal. Since the mirror makes an angle of $60^\circ$ with the vertical, its normal makes an angle of $30^\circ$ with the vertical. Thus, the component of the insect’s velocity along the normal is

$v_{\text{normal}} = v_{\text{max}} \cos 30^\circ = A\sqrt{\frac{K}{M}} \times \frac{\sqrt{3}}{2}$.

Therefore, the maximum relative speed is

$2\,v_{\text{normal}} = 2\left(A\sqrt{\frac{K}{M}}\frac{\sqrt{3}}{2}\right)=A\sqrt{3}\sqrt{\frac{K}{M}}$.