1.

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

Answer»

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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