Question

A particle of mass 0.5 kg executes SHM. If its period of oscillation is `pi` seconds and total energy is 0.04 J, then the amplitude of oscillation will be
A. 40 cm
B. 20 cm
C. 15 cm
D. 10 cm

Answer 1

Correct Answer - B
`T.E.=(1)/(2)momega^(2)A^(2)`
`A^(2)=(2T.E.)/(momega^(2))=(2xx0.4)/(0.5xx(4pi^(2))/(pi^(2)))`
`A^(2)=0.04 therefore A=2m=20cm`