Question

A simple pendulum of length 2m,has a bob of mass 20g and osillates freely with amplitude 6cm.Find it's period and the potential energy of the Bob at an extreme position.

Answer 1

Answer:

Explanation:

time period of simple pendulum is given by, T = 2π√{l/g}

we also know, angular velocity () is given by,

or,

so, time period can be changed into angular velocity ,

given, g = 10m/s² , l = 1m

so,  rad/s

know, potential energy at extreme position, U =

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

Answer 2

I have left some part for you to solve that is calculation of theta..

Try finding theta

I will tell you the answer also but first try finding it....