Q. 61

Expert-verified
Found in: Page 486

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651

Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity is minimum when speaker 2 is 10 cm behind speaker 1. The intensity increases as speaker 2 is moved forward and first reaches maximum, with amplitude 2a, when it is 30 cm in front of speaker 1. What isa. The wavelength of the sound?b. The phase difference between the two loudspeakers?c. The amplitude of the sound (as a multiple of a) if the speakersare placed side by side?

a) The wavelength of the sound is

b) The phase difference between the two loudspeakers are

c) The amplitude of the sound is

See the step by step solution

Part(a) Step 1: Given information

Distance when the speaker 2 is behind the speaker

Amplitude when speaker 2 is in front of speaker at

Part(a) Step 2: Simplification

Formula used:

Distance between the maximum and minimum sound intensities is as below:

\begin{aligned} \Delta x=& 10 \mathrm{~cm}+30 \mathrm{~cm} \\ =& 40 \mathrm{~cm} \end{aligned}

So, the wavelength of the sound between maximum and minimum intensities is as follows:

\begin{aligned} &x=\frac{\lambda}{2} \\ &\lambda=2 \Delta x \end{aligned}

Calculation:

Substitute $\Delta x=40 \mathrm{~cm}$ so, the wavelength will become as below:

\begin{aligned} & \lambda=2 \Delta x \\ =& 2(40 \mathrm{~cm}) \\ =& 80 \mathrm{~cm} \end{aligned}

Conclusion:

Hence, the wavelength of the sound is $80 \mathrm{~cm}$

Part(b) Step 1: Given information

Distance when the speaker 2 is behind the speaker $1, \Delta x=10 \mathrm{~cm}$

Amplitude when speaker 2 is in front of speaker at $30 \mathrm{~cm}, A=2 a$

Part(b) Step 2: Simplification

Formula Used:

The expression for the phase difference between the two sound waves is as below:

$$\Delta \phi=2 \pi \frac{\Delta x}{\lambda}$$

Calculation:

Substitute $\Delta x=40 \mathrm{~cm}$ and $\lambda=80 \mathrm{~cm}$ so, the phase difference is as follows:

\begin{aligned} & \phi=\frac{2 \pi}{\lambda} \Delta x \\ =& \frac{2 \pi}{80 \mathrm{~cm}} 40 \mathrm{~cm} \\ =& 2.36 \mathrm{rad} \end{aligned}

Conclusion:

Hence, the phase difference between the two waves is $2.36 \mathrm{rad}$.

Part(c) Step 1: Given information

Distance when the speaker 2 is behind the speaker $1, \Delta x=10 \mathrm{~cm}$

Amplitude when speaker 2 is in front of speaker at $30 \mathrm{~cm}, A=2 a$

Part(c) Step 2: Simplification

Formula Used:

When two waves with a phase difference $\phi$ and each having an amplitude $a$ then the resulting wave has an amplitude as below:

$$A=2 a \cos \left(\frac{\Delta \phi}{2}\right)$$

Calculation:

Now, substitute $\phi=2.36 \mathrm{rad}$ so, the amplitude is

\begin{aligned} &A=2 a \cos \left(\frac{\Delta \phi}{2}\right) \\ &=2 a \cos \left(\frac{2.36 \mathrm{rad}}{2}\right) \\ &=0.76 a \end{aligned}

Conclusion:

Hence, the amplitude of sound if the speakers are placed side by side is $0.76 a$.