Physics - SHM Question with Solution | TestHub

Let the spring constant of the original spring be $k$ and the mass suspended be $m$.

The time period of vertical oscillations is given by $T_1=2\pi \sqrt{\frac{m}{k}}$.

When the spring is cut into two equal halves, the length of each half becomes half of the original length. Since the spring constant is inversely proportional to the length of the spring ($k\propto \frac{1}{l}$), the spring constant of each half becomes $2k$.

The time period of vertical oscillations when the same mass is suspended from one of the halves is $T_2=2\pi \sqrt{\frac{m}{2k}}$.

The ratio of the time periods is $\frac{T_2}{T_1}=\frac{2\pi \sqrt{\frac{m}{2k}}}{2\pi \sqrt{\frac{m}{k}}}=\frac{1}{\sqrt{2}}$.

Thus, the ratio $T_2:T_1$ is $1:\sqrt{2}$.

Answer: $1:\sqrt{2}$