A simple pendulum bob takes 0.5 seconds to move from mean position to an extreme end, then find the length of the simple pendulum approximately in m (take g $=$ 9.8 m s $^{- 2}$ )

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Solution

The bob takes 0.5 sec to move from mean position to an extreme end which is equal to $\frac{1}{4}$ th of its time period.
so,T=2 s
Again if $l$ is the length of bob
$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow l = \frac{T^{2} g}{4 π^{2}} = \frac{4 \times 9.8}{4 \times π^{2}} = 1.008 m$

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A simple pendulum bob takes 0.5 seconds to move from mean position to an extreme end, then find the length of the simple pendulum approximately in m (take g = 9.8 m s−2)

The bob takes 0.5 sec to move from mean position to an extreme end which is equal to 14 th of its time period.so,T=2 sAgain if l is the length of bobT=2π√lg⇒l=T2g4π2=4×9.84×π2=1.008m

A simple pendulum bob takes 0.5 seconds to move from mean position to an extreme end, then find the length of the simple pendulum approximately in m (take g $=$ 9.8 m s $^{- 2}$ )

The bob takes 0.5 sec to move from mean position to an extreme end which is equal to $\frac{1}{4}$ th of its time period.
so,T=2 s
Again if $l$ is the length of bob
$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow l = \frac{T^{2} g}{4 π^{2}} = \frac{4 \times 9.8}{4 \times π^{2}} = 1.008 m$