Chapter 23, Problem 62SP

The two sources shown in Fig. 23-4 emit identical beams of sound $\left(\lambda =80\text{ cm}\right)$ toward one another. Each sends out a crest at the same time as the other (the sources are in-phase). Point P is a position of

Fig. 23-4

maximum intensity, that is, loud sound. As one moves from P toward Q, the sound decreases in intensity. (a) How far from P will a sound minimum first be heard? (b) How far from P will a loud sound be heard once again?

To determine

The distance from P at which a minimum sound first be heard if two sources emit identical beams of sound $\left(\lambda =80\text{cm}\right)$ toward each other as shown in Fig. 23-4.

Answer to Problem 62SP

Solution:

$20\text{ cm}$

Explanation of Solution

Given data:

The wavelength of sound is $80\text{ cm}$.

Distance between point $P$ and $S_1$ is $5.00\text{ m}$.

Distance between point $P$ and $S_2$ is $5.00\text{ m}$.

Formula used:

The minimum intensity of sound at any point is written as follows:

$L_1-L_2=\pm \left(n+\frac{1}{2}\right)\lambda$

Here, $n$ is an integer;$n=0,1,2,3,\mathrm{.......}$

Here, $L_1$ is the distance of a point from first source, $L_2$ is the distance of a point from second source, $\lambda$ is the wavelength of sound wave, and $n$ is the order of minima.

Explanation:

Consider that the minimum sound is heard at a distance $x$ from point P.

Consider the expression for minimum intensity at distance $x$ from point P as

$\left(L_1+x\right)-\left(L_2-x\right)=\left(n+\frac{1}{2}\right)\lambda$

Point $P$ is at the center of the source so assuming it to be the central maxima and the minima that is heard at distance $x$ from this point to be of the first order. So, $n=0$ at the mentioned distance for the sound to be minimum.

Substitute $5.00\text{ m}$ for $L_1$, $5.00\text{ m}$ for $L_2$, $0$ for $n$, and $80\text{ cm}$ for $\lambda$

$\begin{array}{c}\left[\left(5.00\text{ m}\right)\left(\frac{100\text{ cm}}{1\text{ m}}\right)+x\right]-\left[\left(5.00\text{ m}\right)\left(\frac{100\text{ cm}}{1\text{ m}}\right)-x\right]=\left(0+\frac{1}{2}\right)\left(80\text{ cm}\right)\\ \text{2}x=40\text{ cm}\\ x=20\text{ cm}\end{array}$

Conclusion:

The distance from point P at which minimum sound is heard is $20\text{ cm}$.

To determine

The distance from P at which a loud sound is again heard if two sources emit identical beams of sound $\left(\lambda =80\text{cm}\right)$ toward each other as shown in Fig. 23-4.

Answer to Problem 62SP

Solution:

$40\text{ cm}$

Explanation of Solution

Given data:

The wavelength of sound is $80\text{ cm}$.

Distance between point $P$ and $S_1$ is $5.00\text{ m}$.

Distance between point $P$ and $S_2$ is $5.00\text{ m}$.

Formula used:

The maximum intensity of sound at any point is written as

$L_1-L_2=\pm n\lambda$

Here, $n$ is an integer;$n=0,1,2,3,\mathrm{.......}$

Here, $L_1$ is the distance of a point from first source, $L_2$ is the distance of a point from second source, $\lambda$ is the wavelength of sound wave, and $n$ is the order of maxima.

Explanation:

Consider that the maximum sound is heard at a distance $y$ from point P.

Write the expression for maximum intensity at distance $y$ from point P:

$\left(L_1+y\right)-\left(L_2-y\right)=n\lambda$

Point $P$ is at the center of the source so assuming it to be the central maxima and the maxima that is heard at distance $y$ from this point to be of the first order. So, $n=1$ at the mentioned distance for the sound to be maximum.

Substitute $5.00\text{ m}$ for $L_1$, $5.00\text{ m}$ for $L_2$, $1$ for $n$, and $80\text{ cm}$ for $\lambda$

$\begin{array}{c}\left[\left(5.00\text{ m}\right)\left(\frac{100\text{ cm}}{1\text{ m}}\right)+y\right]-\left[\left(5.00\text{ m}\right)\left(\frac{100\text{ cm}}{1\text{ m}}\right)-y\right]=\left(1\right)\left(80\text{ cm}\right)\\ \text{2}y=80\text{ cm}\\ y=40\text{ cm}\end{array}$

Conclusion:

Therefore, the distance from point P at which the maximum sound is heard is $40\text{ cm}$.