Standing waves on a wire are described by Eq. (15.28), with \({A_{sw}} = 2.5\;{\rm{mm}}\), \(\omega = 942\;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right. \\} {\rm{s}}}\), and \(k = 0.750\pi \;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{m}}}} \right. \\} {\rm{m}}}\). The left end of the wire is at \(x = 0\). At what distances from the left end are (a) the nodes of the standing wave and (b) the antinodes of the standing wave

The given data can be listed below as,

• The amplitude is, \({A_{sw}} = 2.5\;{\rm{mm}}\).
• The value of k is, \(0.750\pi \;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{m}}}} \right. \\} {\rm{m}}}\).

Whenever the wave's amplitude is at its lowest, a standing wave is said to be at its node. A vibrating node might be, for instance, the two ends of a string. The particle of the standing wave is not vibrating at the node's location.

The equation 15.28 is expressed as,

\(y\left( {x,t} \right) = \left( {{A_{sw}}\sin kx} \right)\sin \omega t\) …(1)

Here \({A_{sw}}\) is the amplitude of simple harmonic, \(\omega \) is the angular speed and \(t\) is the time.

Amplitude of zero is found at nodes.

At node,

\(y = 0\)for all time (t). So,

\(\begin{array}{c}\sin kx = 0\\kx = n\pi \\x = \frac{{n\pi }}{k}\end{array}\)

Substitute \(0.750\pi \;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{m}}}} \right. \\} {\rm{m}}}\) for \(k\) in the above equation.

\(\begin{array}{c}x = \frac{{n\pi }}{{0.750\pi \;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{m}}}} \right. \\} {\rm{m}}}}}\\x = \left( {1.33\;{\rm{m}}} \right)n\end{array}\)

Here, \(n = 0,1,2...\)

Hence the distance of nodes of the standing wave from left end is, \(x = \left( {1.33\;{\rm{m}}} \right)n\) and here \(n = 0,1,2...\).