The displacement of a particle executing SHM is given by
`Y=5 " sin "(4t+(pi)/(3))`
If T is the time period and the mass of the particle is 2g, the kinetic energy of the particle When t=`(T)/(4)` is given by-

Options

A. 0.4 J
B. 0.5 J
C. 3 J
D. 0.3 J

Answer

D

Explanation

`y=5sin(4t+pi//3)`
`KE=(1)/(2)m omega^(2)A^(2)cos^(2)omegat`
`=(1)/(2)momega^(2)A^(2)cos^(2)(4t+(pi)/(3))`
`KE=(1)/(2)xx2xx10^(-3)x(4)^(2)xx(5)^(2)cos^(2)(4xx(T)/(4)+(pi)/(3))`
Since `T=(2pi)/(omega)=(2pi)/(4)=(pi)/(2)`
`implies KE=(1)/(2)xx2xx10^(-3)xx(4)^(2)xx(5)^(2)cos^(2)((pi)/(2)+(pi)/(3))`
`=0.4sin^(2)((pi)/(3))=0.4 xx((sqrt3)/(2))^(2)`
`=0.4xx(3)/(4)=0.3J`.