Simple Harmonic Motion

Given, expansion of spring, $\Delta x=0.4 \mathrm{~m}$

Mass suspended in case $\mathbf{I}\left(m_1\right)$ and case II $\left(m_2\right)$ is $0.6 \mathrm{~kg}$ and $255 \mathrm{~g}$, respectively.

Acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$

Since, force $F=m g=k \Delta x$

$\begin{aligned} & \therefore \text { Spring constant } k=\frac{m g}{\Delta x} \\ & \Rightarrow \quad k=\frac{m_1 g}{\Delta x_1}=\frac{0.6 \times 10}{0.4}=15 \mathrm{~N} / \mathrm{m} \end{aligned}$

and time period, $T=2 \pi \sqrt{\frac{m}{k}}$

$\begin{aligned} T_2 & =2 \pi \sqrt{\frac{m_2}{k}} \\ T_2 & =2 \pi \sqrt{\frac{0.255}{15}} \\ & =2 \pi \sqrt{\frac{255}{15 \times 10^3}} \\ & =2 \pi \sqrt{0.017}=0.82 \mathrm{~s} \end{aligned}$

Given, time period of vibration in air = T

Density of fluid = $\rho$/10

Time period of vibration in fluid = T′

Since, $T = 2\pi \sqrt {{m \over k}} $

and as the value of mass (m) and spring constant ( k) of spring-mass system is same.

$\therefore$ Time period will remain same.