Question

For the given equation representing position, $y=\text{A sin}\omega t+A\sin (\omega t+\frac{2\pi }{3})$, match the column.
$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{(A) Motion}& \text{(p) Is periodic but not SHM}\\ \text{(B) Amplitude}& \text{(q) Is SHM}\\ \text{(C) Initial phase}& \text{(r) A}\\ \text{(D) Maximum velocity}& \text{(s)}\frac{\pi }{3}\\ & \text{(t)}\frac{\omega }{3}\\ & \text{(u) None}\end{array}$

• A. $A\to q,B\to r,C\to s,D\to u$
• B. $A\to q,B\to r,C\to u,D\to t$
• C. $A\to p,B\to r,C\to u,D\to r$
• D. $A\to p,B\to u,c\to u,D\to t$

Answer 1

The correct option is A $A\to q,B\to r,C\to s,D\to u$

Given,
$y=\text{A sin}\omega t+\text{A sin}(\omega t+\frac{2\pi }{3})$
This can also be written as
$y=A[\text{sin}\omega t+\text{sin}(\omega t+\frac{2\pi }{3})]...\left(i\right)$
Using trigonometry formula, $\text{sin C}+\text{sin D}=2\text{sin}\left(\frac{C+D}{2}\right)\text{cos}\left(\frac{C-D}{2}\right)$
Eq.$\left(i\right)$ can be written as:
$y=2\text{A sin}(\omega t+\frac{\pi }{3})\text{cos}(-\frac{\pi }{3})$
$\Rightarrow y=2\text{A sin}(\omega t+\frac{\pi }{3})\text{cos}\left(\frac{\pi }{3}\right)$
$\Rightarrow y=\text{A sin}(\omega t+\frac{\pi }{3})...\left(ii\right)$
Comparing the above equation with $y=\text{A sin}(\omega t+\varphi )$
we can conclude that equation $\left(ii\right)$ represents SHM with amplitude $=A$, initial phase $\left(\varphi \right)=\frac{\pi }{3}.$
Differentating Eq.$\left(ii\right)$ w.r.t ${}^{\prime }{t}^{\prime }$ on both sides we get,
$v=\frac{dy}{dt}=\omega \text{A cos}(\omega t+\frac{\pi }{3})$
Velocity$\left(v\right)$ will attain maximum value when, $A\text{cos}(\omega t+\frac{\pi }{3})=1$
$\therefore v_{max}=\omega A$