Question

Two speakers placed 0.94 m apart produce pure tones in sync with each other at a frequency of 1630 Hz. A microphone can be moved along a line parallel to the line joining the speakers and 9.4 m from it. An intensity maximum is measured a point P0 where the microphone is equidistant from the two speakers. As we move the microphone away from P0 to one side, we find intensity minima and maxima alternately. Take the speed of sound in air to be 344 m/s, and you can assume that the slits are close enough together that the equations that describe the interference pattern of light passing through two slits can be applied here.
Required:
a. What is the distance, in meters, between Po and the first intensity minimum?
b. What is the distance, in meters, between Po and the first intensity maximum?
c. What is the distance, in meters, between Po and the second intensity minimum?
d. What is the distance, in meters, between Po and the second intensity maximum?

Answer 1

Answer:

a. approximately $1.1\; \rm m$ (first minimum.)

b. approximately $2.2\; \rm m$ (first maximum.)

c. approximately $3.4\; \rm m$ (second minimum.)

d. approximately $4.7\; \rm m$ (second maximum.)

Explanation:

Let $d$ represent the separation between the two speakers (the two "slits" based on the assumptions.)

Let $\theta$ represent the angle between:

• the line joining the microphone and the center of the two speakers, and
• the line that goes through the center of the two speakers that is also normal to the line joining the two speakers.

The distance between the microphone and point $P_0$ would thus be $9.4\, \tan(\theta)$ meters.

Based on the assumptions and the equation from Young's double-slit experiment:

$\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}$.

Hence:

$\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right)$.

The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let $\lambda$ denote the wavelength of this wave.

$\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}$.

Calculate the wavelength of this wave based on its frequency and its velocity:

$\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m$.

Calculate $\theta$ for each of these path differences:

$\displaystyle \begin{array}{c|c|c} & \text{Path difference} & \text{approximate of \theta } \\ \cline{1-3}\text{First Minimum} & \lambda / 2 & 0.112 \\ \cline{1-3} \text{First Maximum} & \lambda & 0.226\\\cline{1-3} \text{Second Minimum} & 3\,\lambda / 2 & 0.343\\ \cline{1-3} \text{Second Maximum} & 2\, \lambda & 0.466\end{array}$.

In each of these case, the distance between the microphone and $P_0$ would be $9.4\, \tan(\theta)$. Therefore:

• At the first minimum, the distance from $P_0$ is approximately $1.1\; \rm m$.
• At the first maximum, the distance from $P_0$ is approximately $2.2\; \rm m$.
• At the second minimum, the distance from $P_0$ is approximately $3.4\; \rm m$.
• At the second maximum, the distance from $P_0$ is approximately $4.7\;\rm m$.