Question

A block of mass $1\text{kg}$ is attached to a spring of spring constant $100\text{N/m}$. The block is at the centre of curvature of the concave mirror in equilibrium state. It is displaced by $5\text{mm}$ towards the pole and released. The maximum speed of oscillation of the image of the block is

• A. $6\text{cm/s}$
• B. $4\text{cm/s}$
• C. $5\text{cm/s}$
• D. $7\text{cm/s}$

Answer 1

The correct option is C $5\text{cm/s}$

Given:
Mass of block, $m=1\text{kg}$
Spring constant, $K=100\text{N/m}$
Amplitude of oscillation, $A=5\text{mm}$

So, angular frequency, $\omega =\sqrt{\frac{K}{m}}$

$\Rightarrow \omega =\sqrt{\frac{100}{1}}=10\text{rad/s}$ ...........(1)

And maximum speed of the block will be at mean position (at point $C$).
$V_o=A\omega =5\times {10}^{-3}\times 10=0.05\text{m/s}$

Let speed of image along the principal axis be $V_i$.

$V_i=-m^2\times V_o$

Since, the amplitude of oscillation $\left(5\text{mm}\right)$ is very small compared to the focal length of the mirror $\left(20\text{cm}\right)$, we can assume that the magnification $m$ is roughly constant.

Here, $v=u\approx 2f=-40\text{cm}$

From magnification formula,
$m=\frac{-v}{u}=-\frac{-40}{-40}$

$\Rightarrow m=-1$

Since, the object (block) velocity is maximum at $C$, corresponding image velocity will also be maximum.

Thus, at point $C$,

$V_i=-(-1{)}^2\times 0.05$
$\Rightarrow V_i=-0.05\text{m/s}=-5\text{cm/s}$
[-ve sign shows the direction of speed of image which is opposite to the direction of speed of object.]

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