Question

A vertical spring mass system has the same time period as simple pendulum undergoing small oscillations. Now, both of them are put in an elevator going downwards with an acceleration $5 \,m / s ^{2}$. The ratio of time period of the spring mass system to the time period of the pendulum is (Assume, acceleration due to gravity, $g=10 \,m / s ^{2}$ )

• A. $\sqrt{\frac{3}{2}}$
• B. $\sqrt{\frac{2}{3}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{2}$

Answer 1

Answer: (C)

We know that time period of a spring mass system,
$T_{1}=2 \pi \sqrt{\frac{m}{k}}$
where, $m=$ mass of body and $k=$ force constant of the spring Time period of simple pendulum,
$T_{2}=2 \pi \sqrt{\frac{l}{g}}$
According to the question, initially time period of a spring mass system, $T_{1}=$ time period of simple pendulum, $T_{2}$
$2 \pi \sqrt{\frac{m}{k}} =2 \pi \sqrt{\frac{l}{g}}$
$\sqrt{\frac{m}{k}} =\sqrt{\frac{l}{10}} \,\,\,...(i)\left[\because g=10\, m / s ^{2}\right]$
When both of them are put in an elevator going downwards with an acceleration $5 m / s ^{2}$, then no effect occurs on the time period of spring mass system.
i.e.,
$T_{1}^{'}=T_{1}=2 \pi \sqrt{\frac{m}{k}}$
But time period of simple pendulum changes in elevator and is given by
$T_{2}^{'}=2 \pi \sqrt{\frac{l}{g_{ eff }}}, \text { where } g_{ eff }=g-a=$ effective
acceleration of the pendulum when elevator is accelerating downwards with $a m / s ^{2}$.
$T_{2}^{'}=2 \pi \sqrt{\frac{l}{g-5}}=2 \pi \sqrt{\frac{l}{5}}$
$\therefore \frac{T_{1}^{'}}{T_{2}^{'}}=\frac{2 \pi \sqrt{\frac{m}{k}}}{2 \pi \sqrt{\frac{l}{5}}}=\frac{\sqrt{1 / 10}}{\sqrt{1 / 5}} \,\,\,[$ From Eq. (i) $]$
$\therefore \frac{T_{1}^{'}}{T_{2}^{''}}=\frac{1}{\sqrt{2}}$
Hence, the ratio of time period of the spring mass system to the time period of the pendulum is $\frac{1}{\sqrt{2}}$.