Two sources, \(A\) and \(B\), emit a sound of a certain wavelength. The sound emitted from both sources is detected at a point away from the sources. The sound from source \(\mathrm{A}\) is a distance \(d\) from the observation point, whereas the sound from source \(\mathrm{B}\) has to travel a distance of \(3 \lambda .\) What is the largest value of the wavelength, in terms of \(d\), for the maximum sound intensity to be detected at the observation point? If \(d=10.0 \mathrm{~m}\) and the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\), what is the frequency of the emitted sound?

Short Answer

Expert verified
The largest value of the wavelength for the maximum sound intensity to be detected at the observation point is 2.5 meters, and the frequency of the emitted sound is 136 Hz.

Step by step solution

01

Define the given information

Based on the given information, we are given that source A is a distance \(d\) from the observation point, while source B is a distance of \(3\lambda\) from the observation point. Let's denote the path difference as \(\delta = d - 3\lambda\). We are also given the speed of sound, \(v = 340\,\mathrm{m/s}\).
02

Determine the condition for constructive interference

Constructive interference occurs when the waves from the two sources arrive at the observation point in phase. That is, the path difference \(\delta\) must be an integer multiple of the wavelength \(\lambda\). Mathematically, this can be written as: $$\delta = n\lambda$$ Where \(n\) is an integer.
03

Obtain an expression for the largest wavelength for maximum intensity

We need the largest value of \(\lambda\) to have constructive interference (maximum intensity) at the observation point, so we must take the smallest possible value of \(n\), which is \(n = 1\). We can write the equation as: $$d - 3\lambda = \lambda$$ To obtain an expression for the largest wavelength, we can rearrange the equation as follows: $$\lambda = \frac{d}{4}$$
04

Calculate the largest wavelength

Given that \(d = 10.0\,\mathrm{m}\), substitute this value into the expression for the largest wavelength: $$\lambda_{max} = \frac{d}{4} = \frac{10.0\,\mathrm{m}}{4} = 2.5\,\mathrm{m}$$
05

Calculate the frequency

We can use the speed of sound formula to determine the frequency of the emitted sound: $$f = \frac{v}{\lambda_{max}}$$ Substitute the given values, \(v = 340\,\mathrm{m/s}\) and \(\lambda_{max} = 2.5\,\mathrm{m}\): $$f = \frac{340\,\mathrm{m/s}}{2.5\,\mathrm{m}} = 136\,\mathrm{Hz}$$ The largest value of the wavelength for the maximum sound intensity to be detected at the observation point is \(2.5\,\mathrm{m}\), and the frequency of the emitted sound is \(136\,\mathrm{Hz}\).

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Most popular questions from this chapter

You are driving along a highway at \(30.0 \mathrm{~m} / \mathrm{s}\) when you hear a siren. You look in the rear-view mirror and see a police car approaching you from behind with a constant speed. The frequency of the siren that you hear is \(1300 \mathrm{~Hz}\). Right after the police car passes you, the frequency of the siren that you hear is \(1280 \mathrm{~Hz}\). a) How fast was the police car moving? b) You are so nervous after the police car passes you that you pull off the road and stop. Then you hear another siren, this time from an ambulance approaching from behind. The frequency of its siren that you hear is \(1400 \mathrm{~Hz}\). Once it passes, the frequency is \(1200 \mathrm{~Hz}\). What is the actual frequency of the ambulance's siren?

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