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Question

Question: If different types of pendulum are taken to moon, the time period will change for: A. Simple pendu...

If different types of pendulum are taken to moon, the time period will change for:
A. Simple pendulum
B. Spring pendulum
C. Torsional pendulum
D. All of the above

Explanation

Solution

The time period different pendulums depends on different elements such as gravity, length of pendulum, mass of the pendulum and also mass of the object attached to the pendulum. Considering different formulas of different pendulums and deriving the relation between time period and gravity or mass we can easily solve the problem.

Complete step by step answer:
Now, as per the question given here, we will consider three pendulums i.e. simple pendulum, spring pendulum and torsional pendulum.
Now, formula of simple pendulum is T=2πlgT=2\pi \sqrt{\dfrac{l}{g}} ……………………..(i)
Where, T is time period, l is length of string and g is gravitational acceleration.
Formula for spring pendulum is T=2πmkT=2\pi \sqrt{\dfrac{m}{k}} ……………………(ii)
Where, m is mass of spring and k is spring constant.
And formula for torsional pendulum is T=2πIkT=2\pi \sqrt{\dfrac{I}{k}} ………………………(iii)
Where, I am a moment of inertia and here we will take I=mk2I=m{{k}^{2}} for simplicity. So again, the equation becomes T=2πmkT=2\pi \sqrt{\dfrac{m}{k}}.
Now, from expression (i), taking the length of string as constant, it can be clearly observed that the time period is inversely proportional to gravity. And as we know that gravity on the moon changes compared to earth so the time period will also change.
Now, considering equation (ii) and (iii), the time period in these cases does not depend on gravitational acceleration. Hence option (b) and (c) are not correct.
Therefore, the time period will change for a simple pendulum when taken to the moon.

Hence the correct answer is option (a).

Note:
For solving this type of problem students must know the formulas of all the pendulums. Now, in cases of spring pendulum and torsional pendulum, one can also consider relation between mass and time period, as mass of anything becomes 16th{{\dfrac{1}{6}}^{th}} the mass of earth. So, if a student knows this fact then the problem can be solved by this method also.