Question

(a) If a standing wave on a string is given by

$\mathit{y}\left(t\right)\mathbf{=}\left(3mm\right)\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(5x\right)\mathbf{}\mathit{c}\mathit{o}\mathit{s}\left(4t\right)$

is there a node or an antinode of the oscillations of the string atx = 0? (b) If the standing wave is given by

$\mathit{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{3}\mathbf{mm}\mathbf{\right)}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{5}\mathbf{x}\mathbf{+}\mathbf{p}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathbf{4}\mathbf{t}\mathbf{\right)}$

is there a node or an antinode at x = 0?

Short answer

1. The given standing wave has ‘node’ at x = 0
2. The given standing wave has ‘anti node’ at x = 0

Step-by-step solution

Given

1. The equation of a standing wave on a string is, $y\text{'}\left(\mathrm{t}\right)=\left(3\mathrm{mm}\right)sin\left(5\mathrm{x}\right)cos\left(4\mathrm{t}\right)$
2. The equation of a standing wave on a string is,$y\text{'}\left(\mathrm{t}\right)=\left(3\mathrm{mm}\right)sin(5\mathrm{x}+\frac{\mathrm{\tau \tau }}{2})cos\left(4\mathrm{t}\right)$

Determining the concept

Determine the node or antinode using the value of amplitude.

(a) Determining the node or antinode of the oscillations of the string at x=0 if a standing wave is y'(t)=(3mm) sin(5x) cos(4t)

The given equation of a standing wave on a string is,

$y\text{'}\left(\mathrm{t}\right)=\left(3\mathrm{mm}\right)sin\left(5\mathrm{x}\right)cos\left(4\mathrm{t}\right)$

Now, usingin this equation,

$y\text{'}\left(\mathrm{t}\right)=\left(3\mathrm{mm}\right)sin\left(0\right)cos\left(4\mathrm{t}\right)=0$

As the amplitude of the wave becomes zero at this position, a standing wave has node at x=0.

Hence, the given standing wave has ‘node’ at x = 0.

(b) Determining the node or antinode of the oscillations of the string at x=0 if a standing wave is y'(t)=(3mm) sin(5x+ττ2) cos(4t)

The given equation of a standing wave on a string is,

$y\text{'}\left(\mathrm{t}\right)=\left(3\mathrm{mm}\right)sin(5\mathrm{x}+\frac{\mathrm{\tau \tau }}{2})cos\left(4\mathrm{t}\right)$

Now, using x =0 in this equation,

$y\text{'}\left(\mathrm{t}\right)=\left(3\mathrm{mm}\right)sin\left(\frac{\mathrm{\tau \tau }}{2}\right)cos\left(4\mathrm{t}\right)=\left(3mm\right)\cos \left(4t\right)$

This is the maximum amplitude possible for the wave, so at x=0 the given wave has an antinode.

Hence, the given standing wave has ‘antinode’ at x = 0

Therefore, the nodes and antinodes can be determined using the equation of amplitudes.