## 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.