A particle is subjected to two mutually perpendicular Simple Harmonic Motions such that its coordinates at any time t are given by x = A ( t) and y = A ( t) . What is the shape of the particle's trajectory, and what is the angle between its velocity vector and position vector at any instant t > 0 ?

Step-by-step solution

The position of the particle is given by x = A ( t) and y = A ( t) . Squaring and adding the equations, we get: x^2 + y^2 = A^2 ^2( t) + A^2 ^2( t) = A^2 This represents the equation of a circle. Thus, the trajectory of the particle is circular. The position vector is r = A ( t) i + A ( t) j . The velocity vector is obtained by differentiating the position vector with respect to time: v = d r dt = A ( t) i - A ( t) j Taking the dot product of r and v : r v = (A ( t))(A ( t)) + (A ( t))(-A ( t)) = 0 Since the dot pr