Question

Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is
(a) $x(t)=B\sin\bigg \lgroup \frac{2\pi t}{30}\bigg \rgroup$(b) $x(t)=B\cos\bigg \lgroup \frac{\pi t}{15}\bigg \rgroup$(c) $x(t)=B\sin\bigg \lgroup \frac{\pi t}{15}+\frac{\pi t}{2}\bigg \rgroup$(d) $x(t)=B\cos\bigg \lgroup \frac{\pi t}{15}+\frac{\pi t}{2}\bigg \rgroup$

Answer 1

HIII....

I THINK OPTION C IS CORRECT

Answer 2

Answer:

a) Bsin (2πt/30)

Explanation:

Suppose particle P is moving uniformly on a circle of radius A with an angular speed. The two feets of perpendicular Q and R are drawn from P on two diameters.

Let particle P is on the X-axis at t = 0. Radius OP makes an angle with the X-axis at time t, then

x = A cosωt and y = A sinωt

Thus, x and v are the displacements of Q and R from the origin at time = t, which are thus the displacement equations of SHM. This implies that although P is under uniform circular motion, Q and R are performing SHM about O with the same angular speed as that of P.

Thus, Sin ∅ = x/OP = x/B

x = Bsin ∅ = B sin ωt = Bsin (πt/15)

x = Bsin (2πt/30)