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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))` |
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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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