Physics - SHM Question with Solution | TestHub

Let the spring constant be $k$.

When the $500\text{ g}$ mass is removed, the remaining mass attached to the spring is $m_1=100\text{ g}+300\text{ g}=400\text{ g}$.

The time period of oscillation is given by $T_1=2\pi \sqrt{\frac{m_1}{k}}$.

Given $T_1=3\text{ s}$, we have $3=2\pi \sqrt{\frac{400}{k}}$.

When the $300\text{ g}$ mass is also removed, the remaining mass is $m_2=100\text{ g}$.

The new time period is $T_2=2\pi \sqrt{\frac{m_2}{k}}=2\pi \sqrt{\frac{100}{k}}$.

Taking the ratio of the two time periods:

$\frac{T_2}{T_1}=\sqrt{\frac{m_2}{m_1}}=\sqrt{\frac{100}{400}}=\frac{1}{2}$

Therefore, $T_2=\frac{T_1}{2}=\frac{3}{2}=1.5\text{ s}$.

Answer: $1.5$ s