Solution to 2008 Problem 46

Sound will first disappear when sound waves originating from center and from either end of the slit are 180^{\circ} out of phase. This will occur when the path difference for waves originating from the center and from either end of the slit is \lambda/2, where \lambda is the wavelength of the sound wave. This path difference will be \lambda/2 at an angle of 45^{\circ} if the following condition is satisfied
\begin{align*}\frac{0.14 \mbox{ m}}{2} \cdot \sin 45^{\circ} = \lambda/2\end{align*}
The relationship between \lambda, the frequency f, and the speed of sound v, is
\begin{align*}\lambda f = v\end{align*}
So, the condition for destructive interference at 45^{\circ} in terms of the frequency is
\begin{align*}\frac{0.14 \mbox{ m}}{2} \cdot \sin 45^{\circ} = v/2f\end{align*}
Plugging in v=350 \mbox{ m/s} and solving for f gives
\begin{align*}f &= \frac{350 \mbox{ m/s}}{\displaystyle 2 \cdot \frac{0.14 \mbox{ m}}{2} \cdot \frac{\sqrt{2}}{2} } \\&am...
Therefore, answer (D) is correct.

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