Q42E
Question
The wave function of a standing wave is \(y\left( {x,t} \right) = 4.44\;{\rm{mm}}\sin \left[ {\left( {32.5\;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{m}}}} \right. \\} {\rm{m}}}} \right)x} \right]\sin \left[ {\left( {754\;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right. \\} {\rm{s}}}} \right)t} \right]\). For the two traveling waves that make up this standing wave, find the (a) amplitude;
(b) wavelength;
(c) frequency;
(d) wave speed;
(e) wave functions.
(f) From the information given, can you determine which harmonic this is? Explain.
Step-by-Step Solution
Verified(a) The amplitude is, \(2.22\;{\rm{mm}}\).
The given data can be listed below as,
The equation is,
\(\left( {x,t} \right) = 4.44\;{\rm{mm}}\sin \left[ {\left( {32.5\;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{m}}}} \right. \\} {\rm{m}}}} \right)x} \right]\sin \left[ {\left( {754\;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right. \\} {\rm{s}}}} \right)t} \right]\) …(1)
The maximum deviation from equilibrium of a point on a vibrating body or wave in terms of displacement or distance travelled. It is equivalent to the vibration path's half-length.
The standard equation of the standing wave is expressed as,
\(y\left( {x,t} \right) = \left( {{A_{sw}}\sin kx} \right)\sin \omega t\) …(2)
Here \({A_{sw}}\) is the amplitude of simple harmonic and \(\omega \) is the angular frequency.
But,
\({A_{sw}}=2A\) …(3)
Compare both the equation (1) and (2),
\({A_{sw}} = 4.44\;{\rm{mm}}\)
Substitute the value of \({A_{sw}}\) in the equation (3).
\(\begin{array}{c}2A = 4.44\;{\rm{mm}}\\A = \frac{{4.44\;{\rm{mm}}}}{2}\\A = 2.22\;{\rm{mm}}\end{array}\)
Hence the amplitude is, \(2.22\;{\rm{mm}}\).