Chapter 27, Problem 19P

To determine

The angles measured from normal to line joining speakers at which the observer can hear maximum and minimum intensity of sound.

Answer to Problem 19P

Maxima are obtained at $0°,29.1°,76.3°$ and minimum sound is obtained at $14.1°,46.8°$.

Explanation of Solution

Write the equation to find the wavelength of wave produced by oscillator in speaker.

    $\lambda =\frac{c}{f}$

Here, $\lambda$ is the wavelength, $c$ is the speed of light in vacuum and $f$ is the frequency of wave.

Write the condition for interference maxima.

    $m\lambda =d\sin \theta$

Here, $m$ is the order, $\lambda$ is the order, $d$ is the silt separation and $\theta$ is the angular width of fringe.

Rewrite the expression for $\theta$.

  $\theta ={\sin }^{-1}\left(\frac{m\lambda }{d}\right)$        (I)

Write the condition for interference minima.

    $\left(m+\frac{1}{2}\right)\lambda =d\sin \theta$

Rewrite the expression for $\theta$.

  $\theta ={\sin }^{-1}\left(\left(m+\frac{1}{2}\right)\frac{\lambda }{d}\right)$        (II)

Conclusion:

Substitute $3.0\times {10}^8\text{m}/\text{s}$ for $c$ and $2.00\text{kHz}$ for $f$ in the equation for $\lambda$.

    $\begin{array}{c}\lambda =\frac{3.0\times {10}^8\text{m}/\text{s}}{2.00\text{kHz}\left(\frac{{10}^3\text{Hz}}{1\text{kHz}}\right)}\\ =0.170\text{m}\end{array}$

Substitute $0$ for $m$, $0.170\text{m}$ for $\lambda$ and $35.0\text{cm}$ for $d$ in equation (I) to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(0\right)\left(0.170\text{m}\right)}{35.0\text{cm}\left(\frac{\text{1m}}{\text{100}\text{cm}}\right)}\right)\\ =0°\end{array}$

Substitute $1$ for $m$, $0.170\text{m}$ for $\lambda$ and $35.0\text{cm}$ for $d$ in equation (I) to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(1\right)\left(0.170\text{m}\right)}{35.0\text{cm}\left(\frac{\text{1m}}{\text{100}\text{cm}}\right)}\right)\\ =29.1°\end{array}$

Substitute $2$ for $m$, $0.170\text{m}$ for $\lambda$ and $35.0\text{cm}$ for $d$ in equation (I) to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(2\right)\left(0.170\text{m}\right)}{35.0\text{cm}\left(\frac{\text{1m}}{\text{100}\text{cm}}\right)}\right)\\ =76.3°\end{array}$

Substitute $3$ for $m$, $0.170\text{m}$ for $\lambda$ and $35.0\text{cm}$ for $d$ in equation (I) to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(3\right)\left(0.170\text{m}\right)}{35.0\text{cm}\left(\frac{\text{1m}}{\text{100}\text{cm}}\right)}\right)\\ =\text{no}\text{solution}\end{array}$

Substitute $0$ for $m$, $0.170\text{m}$ for $\lambda$ and $35.0\text{cm}$ for $d$ in equation (II) to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(0+\frac{1}{2}\right)\left(0.170\text{m}\right)}{35.0\text{cm}\left(\frac{\text{1m}}{\text{100}\text{cm}}\right)}\right)\\ =14.1°\end{array}$

Substitute $1$ for $m$, $0.170\text{m}$ for $\lambda$ and $35.0\text{cm}$ for $d$ in equation (II) to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(1+\frac{1}{2}\right)\left(0.170\text{m}\right)}{35.0\text{cm}\left(\frac{\text{1m}}{\text{100}\text{cm}}\right)}\right)\\ =46.8°\end{array}$

Substitute $2$ for $m$, $0.170\text{m}$ for $\lambda$ and $35.0\text{cm}$ for $d$ in equation (I) to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(2+\frac{1}{2}\right)\left(0.170\text{m}\right)}{35.0\text{cm}\left(\frac{\text{1m}}{\text{100}\text{cm}}\right)}\right)\\ =\text{no}\text{solution}\end{array}$

Thus, the maxima are obtained at $0°,29.1°,76.3°$ and minimum sound is obtained at $14.1°,46.8°$

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