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?

The amlitude of a particle executes SHM is 2 cm and the force acting a

1.	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?
Answer» SOLUTION :`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))`.