A particle is performing SHM with amplitude 2 cm. At what distance from the equilibrium position is its energy half potential ? What is the maximum velocity of the particle if the frequency of its oscillation is 50 Hz.

Explanation

Data : A =2 cm , PE = KE = `1/2 TE, f = 50 Hz`
` (i) PE = 1/2 kx^(2) = , TE=1/2 kA^(2)`
PE = `1/2` TE
` 1/2 kx^(2) = 1/2(1/2 kA^(2)) " " therefore x^(2) = 1/2 A^(2)`
This is the required distance.
(ii) ` omega = 2pif = 2pi(50) = 100 pi ` rad/s
`v_(max) = omegaA= (100pi)( 2xx 10^(-2)) = 2pi m//s`
AThe particle has equal potenial ank kinetic energies when its displacement is `+-sqrt2 ` cm. and its maximum velocity is ` 2 pi` m/s.