Question

The accelration due to gravity on the surface of moon is 1.7 m/s^2. What is the time period of a simple pendulum on the surface of moon, if its time period on the surface of earth is 3.5 seconds? (g on the surface of earth is 9.8 m/s^2)

Answer 1

Answer:

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Answer 2

$\sf \red{Given}$

$\sf{Gravity \: of \: earth \: (g_{e}) = 9.8 \: m/ s^2}$

$\sf{Gravity \: of \: moon \: (g_{m}) = \: 1.7 m/ s^2}$

$\sf{Time \: period \: on \: earth \: (T_{e}) = 3.5 \: s}$

$\sf \red{We \: have \: to \: find : }$

$\sf{Time \: period \: on \: moon \: (T_{m}) = \: ?}$

$\sf \red{We \: know : }$

$\boxed{ \boxed{T \: = \: 2 \pi \: \sqrt{ \frac{l}{g} } }}$

$\implies \: T \propto \: \frac{1}{ \sqrt{g} }$

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

$\implies \: T_{m} = \bigg( \sqrt{ \frac{g_{e}}{g_{m}} } \: \bigg)T_{e}$

$\sf \red{Putting \: the \: values, \: we \: get : }$

$\sf{ \implies \: T_{m} = \bigg( \sqrt{ \frac{9.8}{1.7} } \: \bigg) \times 3.5}$

$\boxed{\boxed{T_{m} \: = 8.4 \: s}}$