A particle of mass 10g is describing SHM along a straight line with a period of 4s and amplitude 12 cm. What is the kinetic energy when it is (i) 3cm (ii) 6 cm from its equilibrium position. How do you account for the difference between the two values ? `[use pi^(2)=10]`

Explanation

Here, `m=10g, T=4s, A=12cm.`
Case (i), `y=3cm, omega=(2pi)/(T)=(2pi)/(4)=(pi)/(2)rad//s`
velocity, `v=omegasqrt(A^(2)-y^(2))=(pi)/(2)sqrt(144-9)`
`=(pi)/(2)sqrt(135)cm//s`
`KE=(1)/(2)mv^(2)=(1)/(2)xx10xx((pi^(2))/(4)xx135)`
`=168.5 erg`
Case (ii), `y=6cm`
`v=omegasqrt(144-36)=(pi)/(2)sqrt(108)cm//s`
`KE=(1)/(2)mv^(2)=(1)/(2)xx10xx((pi^(2))/(4)xx108)=1350erg`
It means the KE of particle decreases when it moves from `y=3cm ` to `6 cm.` Thi is due to increase in its potential energy.