question_answer

Two simple harmonic motions given by \[x=A\sin (\omega t+\delta )\] and \[y=A\,\sin \left( \omega t+\delta +\frac{\pi }{2} \right)\] act on a particle simultaneously; then the motion of particle will be:                           [AIPMT 2000]

A)  circular anti-clockwise

B)       circular clockwise

C)       elliptical anti-clockwise

D)  elliptical clockwise

Correct Answer: B

Solution :

Two simple harmonic motions be written as

\[x=A\sin (\omega t+\delta )\]                  (i)

and       \[y=A\,\,\sin \,\left( \omega t+\delta +\frac{\pi }{2} \right)\]

or          \[y=A\cos (\omega t+\delta )\]                (ii)

Squaring and adding Eqs. (i) and (ii) we obtain

\[{{x}^{2}}+{{y}^{2}}={{A}^{2}}\,[\sin (\omega t+\delta )+{{\cos }^{2}}(\omega t+\delta )]\]

or         \[{{x}^{2}}+{{y}^{2}}={{A}^{2}}\]          \[(\because {{\sin }^{2}}\,\theta +{{\cos }^{2}}\,\theta =1)\]

This is the equation of a circle.

At         \[\,\omega t+\delta =0\,;\,x=0,\,y=A\]

At         \[\omega t+\delta =\frac{\pi }{2}\,;\,x=A,\,y=0\]

At         \[\omega t+\delta =\pi \,\,\,;\,x=0,\,y=-A\]

At         \[\omega t+\delta =\frac{3\pi }{2}\,\,\,;\,x=-A,\,y=0\]

At         \[\omega t+\delta =2\pi \,\,\,;\,x=0,\,y=A\]

Thus, it is obvious that motion of particle is traversed in clockwise direction.