Question

The spring shown in figure is unstretched when a man starts pulling the block. The mass of the block is $m$. If the man exerts a constant force $F$.
The time period of the motion of the block is

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

Answer 1

The correct option is C $2\pi \sqrt{\frac{m}{k}}$

Here the motion of the block will be simple harmonic. Thus, $F=-kx$ where $x$ be the displacement of block from equilibrium position.

So, $m\frac{d^{2}x}{d{t}^2}=-kx$ or $\frac{d^{2}x}{d{t}^2}=-\left(k/m\right)x$

Comparing the above equation with general equation of SHM, $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$, we get

${\omega }^2=k/m$

or $(2\pi /T{)}^2=k/m$

So, time period $T=2\pi \sqrt{\frac{m}{k}}$