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.

Solution to 2008 Problem 46