1.

Attempt any SIX : A simple pendulum of length 1 m and mass 10 g oscillates freely with amplitude 2 cm. Find its potential energy (P.E.) at the extreme position. `(g = 9.8 m//s^(2))`

Answer» Given : `L = 1m`
`m = 10 g = 10 xx 10^(-3)g`
`a= 2 cm`
`= 2 xx 10^(-2) m`
`t = 2pisqrt((l)/(g))`,
`omega = (2pi)/(T)`
`= (2pi)/(2pisqrt(l/g))`
`= sqrt(g/l) "…"(i)`
At extreme position
P.E. `= 1/2 momega^(2) a^(2)`
`= 1/2 (mg)/(l) a^(2)`
[using equation (i)]
`= 1/2 xx (10 xx 10^(-3) xx 9.8 xx (2 xx 10^(-2))^(2))/(1)`
`= (9.8 xx 4)/(2) xx 10^(-6) J`
`= 9.8 xx 2 xx 10^(-6)`J
`= 19.6 xx 10^(-6) J`
`:. P.E. = 1.96 xx 10^(-5) J`
`:.` Potential energy of S.H.M = `1.96 xx 10^(-5) J`


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