Derive the differential equation for SHM
Let it be µx, where µ is a constant called, the intensity of force. Also on account of a centre of attraction at O, the acceleration of P is towards O i.e., in the direction of x decreasing. Therefore the equation of motion of P is
(d2 x)/(dt2 )=-?x,
where the negative sign has been taken because the force acting on P is towards O i.e., in the direction of x decreasing. The equation (1) gives the acceleration of the particle at any position.
Multiplying both sides of (1) by 2 dx / dt, we get
2 dx/dt (d2 y)/(dt2 )=-2?x dx/dt
Integrating with respect to t, we get v2 = (dx/dt)2 = µx2 + C.
where C is a constant of integration and v is the velocity at P.
Initially at the point A, x = a and v = 0; therefore C = µa2.
Thus, we have
v2=(dx/dt)2= -?x2+?a2 or v2=?(a2-x2 ), (2)