Question

A string of length $L$ is carrying a standing wave of amplitude $A$. Choose the correct options.

• A. $y=A\sin \frac{\pi x}{L}\sin \omega t$, has $2$ nodes.
• B. $y=A\cos \frac{\pi x}{L}\sin \omega t$, has $2$ antinodes.
• C. $y=A\sin \frac{2\pi x}{L}\sin \omega t$, has $2$ antinodes.
• D. $y=A\cos \frac{2\pi x}{L}\sin \omega t$, has $2$ nodes.

Answer 1

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

$y=A\cos \frac{2\pi x}{L}\sin \omega t$, has $2$ nodes.

Comparing given wave equation with standard equation of the standing wave,

$y=2A\sin \left(\frac{n\pi x}{L}\right)\cos \omega t$

(or)

$y=2A\cos \left(\frac{n\pi x}{L}\right)\sin \omega t$

Here $n$ is the number of loops.

Option $\left(A\right)$: $n=1$

For node, $y=0$,

$\Rightarrow$ Nodes will be present at $x=0,L$

$\therefore$ Number of nodes $=2$


Option $\left(B\right):$ $n=1$

$A^{\prime }=\left[A\cos \frac{\pi x}{L}\right]$ should be maximum for antinode.

$\Rightarrow \frac{\pi x}{L}=0,\pi$

$\Rightarrow$ Antinodes will be present at $x=0,x=L$

$\therefore$ Number of antinodes $=2$

Option $\left(C\right):$ $n=2$

For antinode,

$A\sin \frac{2\pi x}{L}$ should be maximum

$\Rightarrow \frac{2\pi x}{L}=+\frac{\pi }{2},\frac{3\pi }{2}$

$\Rightarrow x=\frac{L}{4},\frac{3L}{4}$

Number of antinodes $=2$

Option $\left(D\right):$ $n=2$

For node,

$\cos \frac{2\pi x}{L}$ should be minimum

$\Rightarrow \frac{2\pi x}{L}=\frac{\pi }{2},\frac{3\pi }{2}$

$x=\frac{L}{4},\frac{3L}{4}$

$\therefore$ number of nodes $=2$

Hence, all the given options are correct.

Why this question? This question gives you clear view about standing wave equation and concept of node and antinode.