A simple pendulum of length ' \(\ell\) ' and time period ' \(\mathrm{T}\) ' on earth is taken onto the surface of moon. How should the length of the simple pendulum be changed on the moon such that the time period is constant. (Take \(\left.g_{e}=6 g_{i n}\right)\)

Short Answer

Expert verified
Answer: \(\ell_{in} = \frac{\ell}{6}\)

Step by step solution

01

Write the formula for the time period of a simple pendulum on Earth

The formula for the time period of a simple pendulum is \(T=2\pi\sqrt{\frac{\ell}{g}}\). On Earth, we can write this as \(T=2\pi\sqrt{\frac{\ell}{g_e}}\).
02

Write the formula for the time period of a simple pendulum on the Moon

Using the same formula but considering the gravitational acceleration on the Moon, which is \(g_{in} = \frac{g_e}{6}\), we can write the formula for the time period on the Moon as \(T=2\pi\sqrt{\frac{\ell_{in}}{g_{in}}}\).
03

Make the time periods equal

Since we want the time period to remain constant, we can set the two time period expressions equal to each other: $$2\pi\sqrt{\frac{\ell}{g_e}}=2\pi\sqrt{\frac{\ell_{in}}{g_{in}}}$$
04

Solve for the new length \(\ell_{in}\)

Divide both sides by \(2\pi\) and square both sides to eliminate the square root, we get: $$\frac{\ell}{g_e}=\frac{\ell_{in}}{g_{in}}$$ Now, plug in the values of \(g_e\) and \(g_{in}\) which is \(g_{in} = \frac{g_e}{6}\), and solve for \(\ell_{in}\): $$\ell_{in}=\frac{\ell}{g_e} \times \frac{g_e}{6} = \frac{\ell}{6}$$
05

Final answer

Therefore, in order for the time period to remain constant, the length of the simple pendulum on the Moon must be changed to \(\ell_{in} = \frac{\ell}{6}\).

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