Question

The period of a simple pendulum is 8 s. If the same pendulum experiment is repeated on a plant of g = 1/4 gE, the new period is

Answer 1

Answer:

The new period is 16 s

Explanation:

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g} }$

Where $T$ is the period

$L$ is the length of the string

$g$ is the acceleration due to gravity

2π is constant

For the first experiment,

$T =8 s$

$g = gE$

Then,

$8 = 2\pi \sqrt{\frac{L}{gE} }$

$8\sqrt{gE} = 2\pi \sqrt{L}$ ....... (1)

For the second experiment

$T = T_{2}$

$g = \frac{1}{4} gE$

Hence,

$T=2\pi \sqrt{\frac{L}{g} }$ becomes

$T_{2} =2\pi \sqrt{\frac{L}{\frac{1}{4}gE } }$

Then,

$T_{2} \sqrt{\frac{1}{4}gE } = 2\pi\sqrt{L}$ ....... (2)

Since 2π is constant and

The same pendulum is used for the second experiment, then $L$ is also constant.

∴ $2\pi \sqrt{L}$ is constant for both experiment.

Hence, we can equate equations (1) and (2) such that

$8\sqrt{gE} = T_{2} \sqrt{\frac{1}{4}gE }$

Then,

$\frac{8\sqrt{gE} }{\sqrt{\frac{1}{4}gE } } = T_{2} \\\frac{8\sqrt{gE} }{\frac{1}{2}\sqrt{gE} } } } = T_{2} \\ 16 = T_{2} \\T_{2} = 16 s$

Hence, the new period is 16 s