SHM IN REAL CASES#!#CHAMAK JAEGA SHM HAMESHA K LIYE#!#VISUALISATION

In a S.H.M. with amplitude 'A', what is the ratio of K.E. and P.E. at (A)/(2) distance from the mean position ?

For a particle executing S.H.M., the kinetic energy K is given K = K_(0) cos ^(2)omega t . The maximum value of potential energy is:

The total energy of a particle executing SHM is E. Its kinetic energy is K at the mean position. Then always

Passage IV) Angular frequency in SHM is given by omega=sqrt(k/m) . Maximum acceleration in SHM is omega^(2) A and maximum value of friction between two bodies in contact is muN , where N is the normal reaction between the bodies. Now the value of k, the force constant is increased, then the maximum amplitude calcualted in above question will

A particle executes a S.H.M. Its P.E., K.E. and total energy are measured as function of displacement of x. Then its

A particle executes SHM on a line 8 cm long. Its K.E. and P.E. will be equal when its distance from the mean position is :-