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  • 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 figure shows the circular motion of a particle. The radius of the circle, the period, the sense of revolution and the initial position are indicated in the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is

$$
(a) x(t)=B \sin \left(\frac{2 \pi t}{30}\right)
$$

(b) $x(t)=B \sin \left(\frac{\pi t}{15}\right)$
(c) $x(t)=B \sin \left(\frac{\pi t}{15}+\frac{\pi}{2}\right)$
$(d) x(t)=B \cos \left(\frac{\pi t}{15}+\frac{\pi}{2}\right)$

Answers (1)

Given: T=30s, OQ=B.

The projection of the radius vector on the diameter of the circle when a particle is moving with uniform angular velocity (ω) on a circle of reference is SHM. Let the particle go from P to Q in time t. Then ∠POQ=ωt=∠OQP. The projection of radius OQ on the x-axis will be OR=x(t) say.
image

$$
\begin{aligned}
&\text { In the given triangle } \triangle O Q R \text {, }\\
&\begin{aligned}
& \sin \omega t=\frac{x(t)}{B} \\
& \text { or, } x(t)=B \sin \omega t=B \sin \frac{2 \pi t}{T}=B \sin \frac{2 \pi t}{30}
\end{aligned}
\end{aligned}
$$

Posted by

infoexpert24

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