Question
Sat December 08, 2012 By:

# Derive the differential equation for SHM

Mon December 10, 2012
Let O be the centre of force taken as origin. Suppose the particle starts from rest from the point A where OA = a. It begins to move towards the centre of attraction O. Let P be the position of the particle after time t, where OP = x. By the addition of S.H.M. the magnitude of acceleration at P is proportional to x.
Let it be Âµx, where Âµ 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 = Â–Âµx2 + 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 = Âµa2.

Thus, we have

v2=(dx/dt)2= -?x2+?a2 or v2=?(a2-x2 ), Â…(2)
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