Question

A block whose mass is $1kg$ is fastened to a spring. The spring has a spring constant of $50{Nm}^{-1}$. The block is pulled to a distance of $x=10cm$ from its equilibrium position at $x=0$ $cm$ on a frictionless surface from rest at $t=0$. The kinetic energy of the block when it is $5cm$ away from the mean position is

• A. $0.15J$
• B. $0.12J$
• C. $0.19J$
• D. $0.21J$

Answer 1

Answer: (C)

Angular frequency of S.H.M$=\sqrt{\frac{k}{m}}=5\sqrt{2}$

Here, the body is displaced to maximum at $x=10cm=0.1m$. So, it is the amplitude of the motion velocity of the body performing S.H.M at any position

$x$ then mean position is $v=\omega \sqrt{a^2-x^2}$

Here $a=0.1m$

$x=5cm=0.05m$

$\Rightarrow v=5\sqrt{2}\left(\sqrt{0.01-0.0005}\right)$

$v=0.61m/s$

Kinetic energy at this point$=\frac{1}{2}m{v}^2=\frac{1}{2}\times 1\times {\left(0.61\right)}^2\simeq 0.19J$