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 $s$ ).

0.5

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1.5

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Solution

The correct option is C 1
If 'T' be the time period of simple pendulum,it will take time $\frac{T}{4}$ to go extreme position from its mean position.So here
$\frac{T}{4} = 0.5$
$\Rightarrow T = 2 s$
Now the length $^{'} l^{'}$ of simple pendulum related with time period 'T' as
$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow l = \frac{T^{2} g}{4 π^{2}} \Rightarrow l = 0.99 m \approx 1 m$
Here $g = 9.8 \frac{m}{s^{2}} T = 2 s$

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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).

The correct option is C 1If 'T' be the time period of simple pendulum,it will take time T4 to go extreme position from its mean position.So hereT4=0.5⇒T=2sNow the length ′l′ of simple pendulum related with time period 'T' asT=2π√lg⇒l=T2g4π2⇒l=0.99m≈1mHere g=9.8ms2T=2s

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 $s$ ).