A simple pendulum of the length 1 m has the mass 10g and oscillates freely with amplitude of 5 cm . calculate its potential energy at extreme postion
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Answered by
23
time period of simple pendulum is given by, T = 2π√{l/g}
we also know, angular velocity (\omegaω ) is given by, \omega=\frac{2\pi}{T}ω=
T
2π
or, T=\frac{2\pi}{\omega}T=
ω
2π
so, time period can be changed into angular velocity , \omega=\sqrt{\frac{g}{l}}ω=
l
g
given, g = 10m/s² , l = 1m
so, \omega=\sqrt{10}ω=
10
rad/s
know, potential energy at extreme position, U = \frac{1}{2}m\omega^2A^2
2
1
mω
2
A
2
given, m = 10g = 10^-2 kg, A = 2cm = 2 × 10^-2
so, U = 1/2 × 10^-2 × (√10)² × (2 × 10^-2)²
= 1/2 × 10^-2 × 10 × 4 × 10^-4 J
= 20 × 10^-6 J
= 2 × 10^-5 J
hence, answer is 2 × 10^-5 J
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