An object of mass 10 kg executes SHM. Its time period and amplitude are 2s and 10 m respectively. Find its kinetic energy when it is at a distance of (i) 2 m and (ii) 5 m respectively from its position of equilibrium. Justify the two different results for (i) and (ii).

Explanation

`T=2s, i.e., omega=(2pi)/T=(2pi)/2=pi"rad"*s^(-1)`
Kinetic energy, `K=1/2momega^(2)(A^(2)-x^(2))`
(i) When x = 2m,
`K=1/2xx10xx(pi)^(2)xx{(10)^(2)-(2)^(2)}`
`=5xx(3.14)^(2)xx96=4732.6J`.
(ii) When x = 5 m,
`K=1/2xx10xx(pi)^(2)xx{(10)^(2)-(5)^(2)}`
`=5xx(3.14)^(2)xx75=3697.35J`.
The velocity of a particle executaing SHM becomes maximum at its mean position. As the displacement increases, i.e., as it moves away from the equilibrium position, its velocity decreases and so the kinetic energy also decreases.