Physics - SHM Question with Solution | TestHub

Let the block be pushed down by $x$ from its equilibrium position.

Let the extensions in the springs be $y_1$ and $y_2$ as shown in the figure.

So, the restoring force on the block is $F=K_2{y}_2$.

The force in the left spring is $2F=K_1{y}_1$.

$$2F=K_1{y}_1$$

$$⟹2\times \frac{F}{K_2}+\frac{2F}{K_1}=x$$

$$⟹F=\frac{K_1{K}_2}{2(K_1+K_2)}x=m{\omega }^{2}x$$

So, $\omega =\sqrt{\frac{K_1{K}_2}{2m(K_1+K_2)}}$.

$$⟹T=2\pi \sqrt{\frac{2m(K_1+K_2)}{K_1{K}_2}}$$

$$T=2\pi \sqrt{\frac{2m}{K_{eff}}}$$

$$\therefore n=12$$