Maximum kinetic energy of a particle of mass 1 kg in SHM is 8 J. Time period of SHM is 4 s. Maximum potential energy during the motion is 10 J. Then

Options

A. amplitude of oscillations is approximately 3.53 m
B. minimum potential energy of the particle is 4 J
C. maximum acceleration of the particle is approximately 6.3 `ms^(-2)`
D. minimum kinetic energy of the particle is 2 J

Answer

C

Explanation

Maximum kinetic energy = energy of oscillation in SHM
`therefore" "8=(1)/(2)kA^(2)`
`therefore" "kA^(2)=16" "...(i)`
Further, `2pi sqrt((m)/(k))=4 rArr K = (pi^(2))/(4)" "...(ii)`
From Eqs. (i) and (ii), we get
`k = 2.4 Nm^(-1)and A = 2.53m`
Maximum acceleration of the particle will be
`a_(max)=omega^(2)A=(k)/(m)A=((25)/(1))(2.53)=6.3 ms^(-2)`