Q. 61

Expert-verifiedFound in: Page 486

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 is

a. 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 speakers

are 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

Distance when the speaker 2 is behind the speaker

Amplitude when speaker 2 is in front of speaker at

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 $80 \mathrm{~cm}$

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 $30 \mathrm{~cm}, A=2 a$

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 $2.36 \mathrm{rad}$.

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 $30 \mathrm{~cm}, A=2 a$

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 $0.76 a$.

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