Question

A block of mass $m$ is connected to a spring of spring constant $k$ as shown in Fig. $8.264$. The frame in which the block is placed is given an acceleration a towards left. Neglect friction between the block and the frame walls. The maximum velocity of the block relative to the frame is

• A. $\sqrt{\frac{m}{k}}$
• B. $\alpha \sqrt{\frac{m}{k}}$
• C. $\alpha \sqrt{\frac{m}{2k}}$
• D. $2\alpha \sqrt{\frac{m}{k}}$

Answer 1

Answer: (B)

$\alpha \sqrt{\frac{m}{k}}$

Solve this question relative to the frame (car) pf reference. For maximum velocity (relative to frame), the block must be in equilibrium position.
Let $x_0$ be the equilibrium elongation in spring,k then
$ma=k{x}_0$
From work energy theorem,
$\frac{m{v}^2}{2}0=-\frac{k{x}_0^2}{2}+ma\times x_0$
Solving the above equation, we get $v=a\sqrt{\frac{m}{k}}$
This question is an application of using the work energy theorem in non-inertial frame of reference.