Question

A refrence particle is moving with uniform angular velocity $\omega$ on a circle of reference of radius $a$ with centre at O. The projection of a uniform circular motion of diameter YY' is simple harmonic motion.Time is noted from the instant, the reference particle is at X. Match the displacement in SHM in column I with the phase angle (in rad) of column II.

Answer 1

The displacement equation for the particle executing $SHM$ can be written as $y=asin(wt+\varphi )$

Now at the instant $t=0$, the particle is at x-axis i.e $y=0$

$\therefore 0=asin\left(\varphi \right);At(t=0)⟹\varphi =0$

Thus, displacement equation becomes $y=asinwt$ where , $\left(wt\right)$ represents the phase.

(A): For $y=\frac{a}{2},\frac{a}{2}=asinwt⟹wt=\frac{\pi }{6}OR\frac{5\pi }{6}$

(B): For $y=\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}}=asinwt⟹wt=\frac{\pi }{4}OR\frac{3\pi }{4}$

(C): For $y=\frac{-a\sqrt{3}}{2},\frac{-a\sqrt{3}}{2}=asinwt⟹wt=\frac{4\pi }{3}OR\frac{5\pi }{3}$

(D): For $y=\frac{-a}{\sqrt{2}},\frac{-a}{\sqrt{2}}=asinwt⟹wt=\frac{5\pi }{4}OR\frac{7\pi }{4}$