Question

A vertical spring mass system is taken on moon. It starts oscillating on the moon. Choose the correct statements.

• A. The time period will become $\frac{T_{\text{Earth }}}{\sqrt{6}}$.
• B. The equilibrium position about which spring-mass system oscillates in vertical direction is $\frac{mg}{6k}$ from the unstretched position.
• C. $k$ on moon decreases.
• D. $k$ on moon increases.

Answer 1

Answer: (B)

If $g$ is the acceleration due to gravity on earth, then acceleration due to gravity on moon is $\frac{g}{6}$.
So, when the spring-mass system is taken to the moon, then the FBD is as shown below.

In equilibrium, $kx=\frac{mg}{6}$
where, $k$ is spring constant and $x$ is the extension in the spring.
$x=\frac{mg}{6k}$
Time period of oscillation for spring-mass system,
$T=2\pi \sqrt{\frac{m}{k}}$
Since, the value of $k$ is governed by the elastic properties of the spring only. Also, mass $m$ remains same everywhere irrespective of its position.
Thus, time period $T$ remains unchanged.