Question

The block is displaced towards right through a distance $x$ and released. Then, speed of the block as it passes through the mean position is

• A. $\sqrt{\frac{k_1-k_2}{m}}x$
• B. $\sqrt{\frac{k_2}{m}}x$
• C. $\sqrt{\frac{k_1}{m}}x$
• D. $\sqrt{\frac{k_1+k_2}{m}}x$

Answer 1

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

As block is displaced towards right through a distance $x$, this compresses the left spring and elongates the right spring by the same amount $x$.


Let the speed of the block when it passes through the mean position be $v_0$. Springs will be at their naural lengths when the block passes through mean position.
Applying law of conservation of energy,
$\text{Loss in spring PE}=\text{Gain in KE of block}$
$\Rightarrow \frac{1}{2}{k}_1{x}^2+\frac{1}{2}{k}_2{x}^2=\frac{1}{2}m{v}_0^2$
$\Rightarrow \frac{(k_1+k_2)x^2}{m}=v_0^2$
$\therefore v_0=\sqrt{\frac{k_1+k_2}{m}}x$
Speed of the block will be $\sqrt{\frac{k_1+k_2}{m}}x$ as it passes through the mean position.