Question

There is a point source of sound placed at (0, h) as shown in figure. The detectors $D_1$ and $D_2$ are placed at positions (D, d/2) and (D, -d/2) respectively. Take h <<< D. The source emitted a sound pulse at a certain time. Assuming velocity of sound in the surrounding medium as V. (Assume that the waves reaching the detectors $D_1, D_2$ are plane progressive waves)

If the source emits continuous waves, and the pressure recorded by the two detectors are superposed at every instant in detector $D_0$ (which is equidistant from $D_1\&D_2$) the resultant pressure amplitude will be maximum if the minimum frequency of the source is

• A. $\frac{VD}{2dh}$
• B. $\frac{2VD}{dh}$
• C. $\frac{3VD}{2dh}$
• D. $\frac{VD}{dh}$

Answer 1

Answer: (D)

$\frac{VD}{dh}$

Answer 2

The source is at $S=(0, h)$. The detectors are at $D_1=(D, d/2)$ and $D_2=(D, -d/2)$. The detector $D_0$ is equidistant from $D_1$ and $D_2$. From the figure, $D_0$ is on the x-axis, and it is at $(D, 0)$ which is the midpoint of the segment connecting $D_1$ and $D_2$. Thus, $D_0$ is equidistant from $D_1$ and $D_2$.

The problem states that the pressure recorded by the two detectors $D_1$ and $D_2$ are superposed at every instant in detector $D_0$. This means we are considering the interference of the sound waves from the source that arrive at $D_1$ and $D_2$. The phase difference between the waves arriving at $D_1$ and $D_2$ determines the resultant amplitude when they are superposed.

The distance from the source $S=(0, h)$ to $D_1=(D, d/2)$ is $r_1 = \sqrt{(D-0)^2 + (d/2 - h)^2} = \sqrt{D^2 + (d/2 - h)^2}$. The distance from the source $S=(0, h)$ to $D_2=(D, -d/2)$ is $r_2 = \sqrt{(D-0)^2 + (-d/2 - h)^2} = \sqrt{D^2 + (d/2 + h)^2}$.

The phase difference between the waves arriving at $D_1$ and $D_2$ is given by $\Delta \phi = k(r_2 - r_1) = \frac{2\pi}{\lambda}(r_2 - r_1)$, where $k = 2\pi/\lambda$ is the wave number and $\lambda$ is the wavelength. The resultant pressure amplitude at $D_0$ will be maximum if the waves arriving at $D_1$ and $D_2$ are in phase, i.e., their phase difference is an integer multiple of $2\pi$. $\Delta \phi = 2n\pi$, where $n$ is an integer. $\frac{2\pi}{\lambda}(r_2 - r_1) = 2n\pi$. $r_2 - r_1 = n\lambda$.

We are given $h \ll D$. Let's approximate the distances $r_1$ and $r_2$ using the binomial expansion $\sqrt{a^2 + x^2} = a\sqrt{1 + (x/a)^2} \approx a(1 + \frac{1}{2}\frac{x^2}{a^2})$ for $|x| \ll a$. $r_1 = \sqrt{D^2 + (d/2 - h)^2} = D \sqrt{1 + \frac{(d/2 - h)^2}{D^2}}$. Since $h \ll D$, and assuming $d$ is also not excessively large compared to $D$, we can use the approximation. $r_1 \approx D \left(1 + \frac{(d/2 - h)^2}{2D^2}\right) = D + \frac{(d/2 - h)^2}{2D} = D + \frac{d^2/4 - dh + h^2}{2D}$. $r_2 = \sqrt{D^2 + (d/2 + h)^2} = D \sqrt{1 + \frac{(d/2 + h)^2}{D^2}}$. $r_2 \approx D \left(1 + \frac{(d/2 + h)^2}{2D^2}\right) = D + \frac{(d/2 + h)^2}{2D} = D + \frac{d^2/4 + dh + h^2}{2D}$.

The path difference is $\Delta r = r_2 - r_1 \approx \left(D + \frac{d^2/4 + dh + h^2}{2D}\right) - \left(D + \frac{d^2/4 - dh + h^2}{2D}\right) = \frac{dh - (-dh)}{2D} = \frac{2dh}{2D} = \frac{dh}{D}$.

The condition for maximum amplitude is $\Delta r = n\lambda$, where $n$ is an integer. $\frac{dh}{D} = n\lambda$. We are looking for the minimum frequency $f$. The relationship between frequency, velocity, and wavelength is $V = f\lambda$, so $\lambda = V/f$. Substituting this into the equation: $\frac{dh}{D} = n \frac{V}{f}$. $f = n \frac{VD}{dh}$.

Since $r_2 = \sqrt{D^2 + (d/2+h)^2}$ and $r_1 = \sqrt{D^2 + (d/2-h)^2}$, and $h>0, d>0$, we have $(d/2+h)^2 > (d/2-h)^2$, so $r_2 > r_1$. Thus, the path difference $r_2 - r_1$ is positive. The condition for constructive interference is $\Delta r = n\lambda$ for $n=0, 1, 2, \dots$. If $n=0$, $\Delta r = 0$, which is generally not the case unless $h=0$ or $d=0$. Since $h$ and $d$ are given as non-zero dimensions in the figure, we consider positive integer values for $n$. The possible frequencies for maximum amplitude are $f = \frac{VD}{dh}, \frac{2VD}{dh}, \frac{3VD}{dh}, \dots$ (corresponding to $n=1, 2, 3, \dots$). The minimum frequency occurs for the smallest positive integer value of $n$, which is $n=1$. So, the minimum frequency is $f_{min} = \frac{VD}{dh}$.