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Question
A simple pendulum of length L has mass m and it oscillates freely with amplitude A. At the extreme position, its potential energy is (g = acceleration due to gravity)
Options
`"mgA"/"L"`
`"mgA"/2"L"`
`"mgA"^2/"L"`
`"mgA"^2/"2L"`
MCQ
Solution
`"mgA"^2/"2L"`
Explanation:
At extreme position, the potential energy of simple pendulum,
`"PE"=1/2"mA"^2omega` ...(i)
where, ω = angular frequency = `(2pi)/"T"`
For simple pendulum, T = `2pisqrt("L"/"g"`
∴ `omega=(2pi)/(2pi)sqrt"g"/"L"=sqrt"g"/"L"` ...`[because"T"=(2pi)/omega]`
On putting value of ω in Eq. (i), we get
`"PE"=1/2"mM"^2(sqrt("g"/"L"))^2`
`="mgA"^2/(2"L")`
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Simple Pendulum
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