Question

Time period of oscillation of a spring is $12\,{\text{s}}$ on earth. What shall be the time period if it is taken to the moon?
A.$6\,{\text{s}}$
B.$12\,{\text{s}}$
C.$36\,{\text{s}}$
D.$72\,{\text{s}}$

Answer 1

Use the formula for time period of oscillation of the spring. This formula gives the relation between the time period of oscillation of the spring, mass of the oscillating object and spring constant of the spring. Check the dependence of the time period of oscillation of the spring on acceleration due to gravity and determine the time period of oscillation on the moon.

Formula used:
The time period $T$ of oscillation of a spring is given by
$T = 2\pi \sqrt {\dfrac{m}{K}}$ …… (1)
Here, $m$ is the mass of the oscillating object and $K$ is spring constant of the spring.

Complete step by step answer:
We have given that the time period of oscillation of the spring on the Earth is $12\,{\text{s}}$. We have to calculate the time period of oscillation of the spring on the moon. The time period of oscillation of the spring on the moon will be different only if the time period of oscillation of the spring depends on acceleration due to gravity as it is the only changing term related with the oscillating spring on the moon.

From equation (1), we can conclude that the time period of oscillation of the spring depends only on the mass of the oscillating object and the spring constant of the spring. Both of these physical quantities remain the same everywhere in the universe. Hence, the time period of oscillation of the spring on the moon will remain the same as that on the earth.

Therefore, the time period of oscillation of spring on the moon will be $12\,{\text{s}}$. Hence, the correct option is B.

Note: The students may get confused and think that the time period of oscillation of the spring depends on acceleration due to gravity of the planet. But the students should keep in mind that the time period of oscillation of a simple pendulum depends on acceleration due to gravity and not of the spring.