Question

a simple pendulum complete 4 vibrations in 8 seconds on the surface of earyh find the time period on the surface of moon where acceleration due to gravity is 1/6 of earth​

Answer 1

please mark my answer as brain list

Answer 2

Answer: Time period on surface of moon is $4.898 s$.

Explanation:

As given $4$ vibrations are completed on surface of earth in time $= 8 s$

$1$ vibration is completed on surface of earth in time $= \frac{8}{4} = 2s$

∴ Time period on surface of earth $= 2s$

Mathematically time period on surface of earth;

$T_{e} = 2\pi \sqrt{\frac{L}{g_{e} } }$.....$(1)$

where $L$ corresponds to pendulum length and $g_{e}$ is the acceleration due to gravity on surface of moon.

Now time period on surface of moon;

$T_{m} = 2\pi \sqrt{\frac{L}{g_{m} } }$ .....$(2)$

On dividing equation $(2)$ by $(1)$

$\frac{T_{m} }{T_{e} } = \sqrt{\frac{g_{e} }{g_{m} } }$

Using $g_{m} = \frac{g_{e} }{6}$

$\frac{T_{m} }{T_{e} } = \sqrt{\frac{6g_{e} }{g_{e} } }= \sqrt{6}$

$T_{m} = \sqrt{6} * T_{e} = \sqrt{6} * 2s = 2.449 * 2s = 4.898 s$

Therefore time period on surface of moon is $4.898 s$.

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