Question

Two waves travelling in opposite directions produce a standing wave. The individual wave functions are given by $y_1=4\sin (3x-2t)\text{cm}$ and $y_2=4\sin (3x+2t)\text{cm}$, where $x$ and $y$ are in $\text{cm}$ then :
(Use $\sin \left(6.9\right)=0.57$ if required.)

• A. Nodes are formed at $x$ values given by $0,\frac{\pi }{3},\frac{2\pi }{3},\pi ,\frac{4\pi }{3},\dots$
• B. The maximum displacement of the motion at $x=2.3\text{cm}$ is $4.63\text{cm}$
• C. Antinodes are formed at $x$ values given by $\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\frac{7\pi }{6},\dots$
• D. The maximum displacement of the motion at $x=2.3\text{cm}$ is $5.32\text{cm}$

Answer 1

Answer: (A), (B), (C)

Antinodes are formed at $x$ values given by $\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\frac{7\pi }{6},\dots$

Given that,
$y_1=4\sin (3x-2t)\text{cm}$
$y_2=4\sin (3x+2t)\text{cm}$

By using superposition of waves, the net displacement
$y=y_1+y_2$
$\Rightarrow y=4\sin (3x-2t)+4\sin (3x+2t)$
$\Rightarrow y=4\left[\sin \right(3x-2t)+\sin (3x+2t\left)\right]$
Applying: $\sin (A-B)+\sin (A+B)=2\sin A\cos B$
$\Rightarrow y=4\left(2\sin 3x\cos 2t\right)$
$\Rightarrow y=8\sin 3x\cos 2t$

Now for maximum displacement, considering maximum value of $\cos 2t=1$, we have:
Amplitude, $A=8\sin \left(3x\right)$
The displacement at $x=2.3\text{cm}$ is
$A^{\prime }=8\sin \left[3\right(2.3\left)\right]$
$\Rightarrow A^{\prime }=8\sin \left(6.9\right)$
$\Rightarrow A^{\prime }=4.63\text{cm}$

Nodes are formed where, amplitude or, intensity $=0$ i.e.,
$I_R=A^2=0$
$\Rightarrow A^2={\sin }^2\left(3x\right)=0$
$\Rightarrow \sin \left(3x\right)=0$
$3x=0,\pi ,2\pi ,3\pi$
$\therefore x=0,\frac{\pi }{3},\frac{2\pi }{3},\pi ,\frac{4\pi }{3}\dots$

Anti-node is always formed between two nodes in a standing waves.
The position of antinodes are
$\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\frac{7\pi }{6}\dots$
So, we see corresponding location of anti-nodes lie between the calculated node locations.

Hence, option (a), (c), (d) are correct.