Question

If variation of position of a particle with time is denoted by $y=A\sin \omega t+A\sin (\omega t+\frac{2\pi }{3})$, then
Match Column I with Column II

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{(A) Type of motion}& \text{(p) SHM}\\ \text{(B) Amplitude}& \text{(q) A}\\ \text{(C) Initial Phase}& \text{(r)}\frac{\pi }{3}\\ \text{(D) Maximum velocity}& \text{(s)}A\omega \\ & \text{(t) Not SHM}\\ & \text{(u)}\frac{2\pi }{3}\end{array}$

• A. $A-s,b-q,c-p,d-r$
• B. $A-s,b-p,c-q,d-r$
• C. $A-p,b-q,c-s,d-r$
• D. $A-p,b-q,c-r,d-s$

Answer 1

The correct option is D $A-p,b-q,c-r,d-s$

As $y=A\sin \omega t+A\sin (\omega t+\frac{2\pi }{3})$
$y=A\sin \omega t+A\sin \omega t\cos \frac{2\pi }{3}+A\cos \omega t\sin \frac{2\pi }{3}$
$y=A\sin \omega t+A\sin \omega t\left(\frac{-1}{2}\right)+A\cos \omega t\left(\frac{\sqrt{3}}{2}\right)$
$y=A\sin \omega t\left(\frac{1}{2}\right)+A\cos \omega t\left(\frac{\sqrt{3}}{2}\right)$
$y=A\sin (\omega t+\frac{\pi }{3})$
$\therefore y=A\sin (\omega t+\varphi )\to S.H.M.$
Amplitude of SHM is $A$
Initial phase is $\varphi =\frac{\pi }{3}$
Maximum velocity is $A\omega$

Why this question ? Key Concept: Superpostion of two collinear SHMs of same frequency can be solved by analytical approach.