Physics - SHM Question with Solution | TestHub

Let the spring constant of each identical spring be $k$.

In the given arrangement, when the mass $m$ is displaced by a distance $x$ from its equilibrium position, one spring is stretched by $x$ and the other is compressed by $x$. Both springs exert a restoring force in the same direction.

The total restoring force is $F_{net}=-kx-kx=-2kx$.

Thus, the effective spring constant of the system is $k_{eff}=2k$ (the springs are in parallel).

The initial frequency of oscillation is:
$F=\frac{1}{2\pi }\sqrt{\frac{k_{eff}}{m}}=\frac{1}{2\pi }\sqrt{\frac{2k}{m}}$

When one spring is removed, the mass is attached to only one spring. The new effective spring constant is $k_{eff}^{\prime }=k$.

The new frequency of oscillation is:
$F^{\prime }=\frac{1}{2\pi }\sqrt{\frac{k_{eff}^{\prime }}{m}}=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

Taking the ratio of the new frequency to the initial frequency:
$\frac{F^{\prime }}{F}=\frac{\frac{1}{2\pi }\sqrt{\frac{k}{m}}}{\frac{1}{2\pi }\sqrt{\frac{2k}{m}}}=\frac{1}{\sqrt{2}}$

$\Rightarrow F^{\prime }=\frac{F}{\sqrt{2}}$

Answer: $F/\sqrt{2}$