Find the natural frequency of the system shown in the f

SOLUTION

Let x  be the displacement of the block from the equilibrium position.
Since the length of the string remains the same, the spring extends by x = (2x).
The energy of the system is
$E=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2=\frac{1}{2}k(2x{)}^2+\frac{1}{2}m{v}^2$
Differentiate with respect to time
$\frac{dE}{dt}-4kx\frac{dx}{dt}+mv\frac{dv}{dt}=0$ (energy conservation)
$\Rightarrow 4kxv+mv\frac{d^{2}x}{d{t}^2}=0$
$\therefore 4kg+m\frac{d^{2}x}{d{t}^2}=0\Rightarrow \frac{d^{2}x}{d{t}^2}+\frac{4kx}{m}=0$
$\therefore T=2\pi \sqrt{\frac{m}{4k}}=\pi \sqrt{\frac{m}{k}}$