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 $5m/s^2$. The ratio of time period of the spring mass system to the time period of the pendulum is (Assume, acceleration due to gravity, $g=10m/s^2$ )

• 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, $T_2$
$2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{l}{g}}$
$\sqrt{\frac{m}{k}}=\sqrt{\frac{l}{10}}...(i)\left[\because g=10m/s^2\right]$
When both of them are put in an elevator going downwards with an acceleration $5m/s^2$, then no effect occurs on the time period of spring mass system.
i.e.,
$T_1^{{}^{\prime }}=T_1=2\pi \sqrt{\frac{m}{k}}$
But time period of simple pendulum changes in elevator and is given by
$T_2^{{}^{\prime }}=2\pi \sqrt{\frac{l}{g_{eff}}},\text{ where }{g}_{eff}=g-a=$ effective
acceleration of the pendulum when elevator is accelerating downwards with $am/s^2$.
$T_2^{{}^{\prime }}=2\pi \sqrt{\frac{l}{g-5}}=2\pi \sqrt{\frac{l}{5}}$
$\therefore \frac{T_1^{{}^{\prime }}}{T_2^{{}^{\prime }}}=\frac{2\pi \sqrt{\frac{m}{k}}}{2\pi \sqrt{\frac{l}{5}}}=\frac{\sqrt{1/10}}{\sqrt{1/5}}[$ From Eq. (i) $]$
$\therefore \frac{T_1^{{}^{\prime }}}{T_2^{{}^{\prime \prime }}}=\frac{1}{\sqrt{2}}$
Hence, the ratio of time period of the spring mass system to the time period of the pendulum is $\frac{1}{\sqrt{2}}$.