Question

(III) Two loudspeakers are placed 3.00 m apart, as shown in Fig. 12–37. They emit 474-Hz sounds, in phase. A microphone is placed 3.20 m distant from a point midway between the two speakers, where an intensity maximum is recorded. (a) How far must the microphone be moved to the right to find the first intensity minimum? (b) Suppose the speakers are reconnected so that the 474-Hz sounds they emit are exactly out of phase. At what positions are the intensity maximum and minimum now?

Figure 12-37

Short answer

1. The distance to which the microphone moves is \(0.426\;{\rm{m}}\).
2. The intensity maxima is at \(0.426\;{\rm{m}}\) to the left or right of the midpoint, and the intensity minima is exactly at the midpoint.

Step-by-step solution

Determination of the wavelength of sound waves

The wavelength of the sound wave can be obtained by taking the ratio of the speed of the sound wave and frequency.

Given information

Given data:

The distance between two loudspeakers is \(d = 3.00\;{\rm{m}}\).

The frequency emitted by the loudspeakers is \(f = 474\;{\rm{Hz}}\).

The distance between the microphone and the loudspeakers is \(l = 3.20\;{\rm{m}}\)

Evaluation of the distance to which the microphone moves

(a)

Consider the microscope is moved at a distance of x from its initial position. So, the new distances are shown in the below figure.

The path difference can be given as:

\({d_1} - {d_2} = \frac{\lambda }{2}\)

From the above figure, the equation for the distance changes with respect to xcan be given as:

\(\begin{aligned}{}\sqrt {{{\left( {\frac{d}{2} + x} \right)}^2} + {l^2}} - \sqrt {{{\left( {\frac{d}{2} - x} \right)}^2} + {l^2}} &= \frac{\lambda }{2}\\\sqrt {{{\left( {\frac{d}{2} + x} \right)}^2} + {l^2}} &= \frac{\lambda }{2} + \sqrt {{{\left( {\frac{d}{2} - x} \right)}^2} + {l^2}} \end{aligned}\)

On squaring on both sides, you get:

\(\begin{aligned}{}{\left( {\frac{d}{2} + x} \right)^2} + {l^2} &= \frac{{{\lambda ^2}}}{4} + \lambda \sqrt {{{\left( {\frac{d}{2} - x} \right)}^2} + {l^2}} + {\left( {\frac{d}{2} - x} \right)^2} + {l^2}\\2dx - \frac{{{\lambda ^2}}}{4} &= \lambda \sqrt {{{\left( {\frac{d}{2} - x} \right)}^2} + {l^2}} \\4{d^2}{x^2} - \left( {4dx} \right)\frac{{{\lambda ^2}}}{4} + \frac{{{\lambda ^4}}}{{16}} &= {\lambda ^2}\left( {{{\left( {\frac{d}{2} - x} \right)}^2} + {l^2}} \right)\\x &= \lambda \sqrt {\frac{{\left( {\frac{1}{4}{d^2} + {l^2} - \frac{{{\lambda ^2}}}{{16}}} \right)}}{{\left( {4{d^2} - {\lambda ^2}} \right)}}} \end{aligned}\) … (1)

As you know \(\lambda = \frac{v}{f}\), substitute the value in equation (1), you get:

\(x = \left( {\frac{v}{f}} \right)\sqrt {\frac{{\left( {\frac{1}{4}{d^2} + {l^2} - \frac{1}{{16}}{{\left( {\frac{v}{f}} \right)}^2}} \right)}}{{\left( {4{d^2} - {{\left( {\frac{v}{f}} \right)}^2}} \right)}}} \) … (2)

On substituting the given numerical values in equation (2), you get:

\(\begin{aligned}{c}x &= \left( {\frac{{343\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}}}{{474\;{\rm{Hz}}}}} \right)\sqrt {\frac{{\left( {\frac{1}{4} \times {{\left( {3\;{\rm{m}}} \right)}^2} + {{\left( {3.20\;{\rm{m}}} \right)}^2} - \frac{1}{{16}}{{\left( {\frac{{343\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}}}{{474\;{\rm{Hz}}}}} \right)}^2}} \right)}}{{\left( {4 \times {{\left( {3\;{\rm{m}}} \right)}^2} - {{\left( {\frac{{343\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}}}{{474\;{\rm{Hz}}}}} \right)}^2}} \right)}}} \\x &= \left( {0.72\;{\rm{m}}} \right) \times \left( {\sqrt {\frac{{12.45}}{{35.48}}} } \right)\\x &= 0.426\;{\rm{m}}\end{aligned}\)

Thus, the distance to which the microphone moves is \(0.426\;{\rm{m}}\).

Evaluation of the position of intensity maxima

(b)

Suppose the speakers are reconnected so that they emit exactly \(474\;{\rm{Hz}}\) the frequency sound out of the phase, the position of maxima and minima are interchanged. Hence, the intensity maxima is at \(0.426\;{\rm{m}}\) to the left or right of the midpoint, and the intensity minima is exactly at the midpoint.