(III) Two loudspeakers are 1.60 m apart. A person stands 3.00 m from one speaker and 3.50 m from the other. (a) What is the lowest frequency at which destructive interference will occur at this point if the speakers are in phase? (b) Calculate two other frequencies that also result in destructive interference at this point (give the next two highest). Let T = 20°C.

For the occurrence of destructive interference, one person must be half-wavelength farther from one speaker than from the other one.

Given data:

The distance between two loudspeakers is \(d = 1.60\;{\rm{m}}\)

The distance between person stands from one speaker is \({d_1} = 3.00\;{\rm{m}}\)

The distance between person stands from the other is \({d_2} = 3.50\;{\rm{m}}\)

(a)

For the two loudspeakers, the combined wavelength will be \(\lambda = 1.0\).

The formula for the lowest frequency at which the destructive interference will occur can be given as:

\(f = \frac{v}{\lambda }\) … (1)

Here, v is the speed of sound at \(20^\circ {\rm{C}}\) \(\left( {343\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\).

After substituting the given values in equation (1), you get:

\(\begin{array}{c}f = \frac{{343\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}}}{{1.0\;{\rm{m}}}}\\f = 343\;{\rm{Hz}}\end{array}\)

Hence, the lowest frequency at which the destructive interference will occur is \(343\;{\rm{Hz}}\).

(b)

Let’s assume that you need to place the person at a distance \(\frac{3}{2}\lambda \) from one speaker and\(\frac{5}{2}\lambda \) from the other speaker. Hence,

The path difference can be calculated as:

\[\begin{array}{c}\Delta x &= \left( {3.50\;{\rm{m}}} \right) - \left( {3.00\;{\rm{m}}} \right)\\\Delta x &= 0.5\;{\rm{m}}\end{array}\]

For one speaker,

\(\begin{array}{c}\Delta x &= \frac{3}{2}\lambda \\0.5\;{\rm{m}} &= 1.5 \times \lambda \\\lambda = 0.33\;{\rm{m}}\end{array}\)

Similarly, for the other speaker,

\(\begin{array}{c}\Delta x &= \frac{5}{2}\lambda \\0.5\;{\rm{m}} &= 2.5 \times \lambda \\\lambda &= 0.2\;{\rm{m}}\end{array}\)

After substituting the given values in equation (1), you get:

\(\begin{array}{c}f &= \frac{{343\;{\rm{m/s}}}}{{\left( {0.33\;{\rm{m}}} \right)}}\\f &= 1039.39\;{\rm{Hz}}\end{array}\)

After substituting the given values in equation (i), you get:

\(\begin{array}{c}f &= \frac{{343\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \ } {\rm{s}}}}}{{\left( {0.2\;{\rm{m}}} \right)}}\\f &= 1715\;{\rm{Hz}}\end{array}\)

Thus, the two other frequencies in destructive interference are \(1039.39\;{\rm{Hz}}\) and \(1715\;{\rm{Hz}}\).