The amlitude of a particle executes SHM is 2 cm and the force acting at extreme position on particle is 4N, then what is force at midway between mean position and extreme point?

Explanation

`F= kA`
At extreme position `F= kA`
`4= kxx 2 xx 10^(-2)`
`therefore k= 200 Nm^(-1)`
Now, at midway point `x= (A)/(2) = (2)/(2)= 1cm`
Force `F = k(1xx 10^(-2))`
`= 200 xx 1xx 10^(-2)= 2N`
Now, maximum acceleration
`a_("max") = omega^(2)A`
`= omega^(2) sqrt(A_(1)^(2) +A_(2)^(2)+ 2A_(1) A_(2))`.