Chapter 14, Problem 36P

A yellow submarine traveling horizontally at 11.0 m/s uses sonar with a frequency of 5.27 × 10³ Hz. A red submarine is in front of the yellow submarine and moving 3.00 m/s relative to the water in the same direction. A crewman in the red submarine observes sound waves (“pings”) from the yellow submarine. Take the speed of sound in seawater as 1 533 m/s. (a) Write Equation 14.12. (b) Which submarine is the source of the sound? (c) Which submarine carries the observer? (d) Does the motion of the observer’s submarine increase or decrease the time between the pressure maxima of the incoming sound waves? How does that affect the observed period? The observed frequency? (e) Should the sign of v₀ be positive or negative? (f) Does the motion of the source submarine increase or decrease the time observed between the pressure maxima? How does this motion affect the observed period? The observed frequency? (g) What sign should be chosen for v_s? (h) Substitute the appropriate numbers and obtain the frequency observed by the crewman on the red submarine.

To determine

The equation 14.12.

Answer to Problem 36P

The equation 14.12 in the text book is,

$f_0=f_s\left(\frac{v+v_0}{v-v_s}\right)$

• $f_s$ is the frequency emitted by the source
• $f_0$ is the frequency detected by the observer
• $v$ is the speed of the wave in the propagating medium
• $v_0$ is the velocity of the observer relative to the medium
• $v_s$ is the source’s velocity relative to the propagating medium

Explanation of Solution

The equation 14.12 in the text book is,

$f_0=f_s\left(\frac{v+v_0}{v-v_s}\right)$

• $f_s$ is the frequency emitted by the source
• $f_0$ is the frequency detected by the observer
• $v$ is the speed of the wave in the propagating medium
• $v_0$ is the velocity of the observer relative to the medium
• $v_s$ is the source’s velocity relative to the propagating medium

Conclusion:

The equation 14.12 in the text book is,

$f_0=f_s\left(\frac{v+v_0}{v-v_s}\right)$

• $f_s$ is the frequency emitted by the source
• $f_0$ is the frequency detected by the observer
• $v$ is the speed of the wave in the propagating medium
• $v_0$ is the velocity of the observer relative to the medium
• $v_s$ is the source’s velocity relative to the propagating medium

To determine

The submarine which is the source of sound.

Answer to Problem 36P

The source of sound is yellow submarine.

Explanation of Solution

The yellow submarine emits the sound.

Thus, the source of sound is the yellow submarine.

Conclusion:

The source of sound is yellow submarine.

To determine

The submarine which is carrying the observer.

Answer to Problem 36P

The red submarine is carrying the observer.

Explanation of Solution

The observer is standing in the red submarine and observing the sound wave coming from the yellow submarine.

Thus, the red submarine is carrying the observer.

Conclusion:

The red submarine is carrying the observer.

To determine

Whether the observers submarine increase or decrease the time between the pressure maxima of the incoming sound waves.

Answer to Problem 36P

The observers submarine increase the time between the pressure maxima of the incoming sound waves.

Explanation of Solution

Since, the submarine carrying the observer is moving; the observer is also moving. The observer is moving away from the observer. So, this will increase the total time observed between the arrivals of successive pressure maxima. This effect will decrease the observed frequency and an increase in the observed period.

Thus, the observers submarine increase the time between the pressure maxima of the incoming sound waves.

Conclusion:

The observers submarine increase the time between the pressure maxima of the incoming sound waves.

To determine

The sign of the velocity of the observer relative to the medium, $v_0$.

Answer to Problem 36P

The sign of the velocity of the observer relative to the medium ( $v_0$) is negative.

Explanation of Solution

Given Info:

$f_0=f_s\left(\frac{v+v_0}{v-v_s}\right)$

• $f_s$ is the frequency emitted by the source
• $f_0$ is the frequency detected by the observer
• $v$ is the speed of the wave in the propagating medium
• $v_0$ is the velocity of the observer relative to the medium
• $v_s$ is the source’s velocity relative to the propagating medium

Since, the observed frequency is decreasing; the sign of the variable $v_0$ should be negative in the above equation.

Thus, the sign of the velocity of the observer relative to the medium ( $v_0$) is negative.

Conclusion:

The sign of the velocity of the observer relative to the medium ( $v_0$) is negative.

To determine

Whether the motion of the source submarine increase or decrease the time between the pressure maxima and how does this motion affect the observed period and the frequency.

Answer to Problem 36P

The time between the arrivals of the successive pressure maxima will decrease. The observed period and the frequency will be increasing.

Explanation of Solution

Given Info:

Since, the source submarine is moving towards the observer; the time between the arrivals of successive pressure maxima will be decreasing. Thus, the observed period will be increasing and the observed frequency will also be increasing.

Thus, the time between the arrivals of the successive pressure maxima will decrease. The observed period and the frequency will be increasing.

Conclusion:

The time between the arrivals of the successive pressure maxima will decrease. The observed period and the frequency will be increasing.

To determine

The sign of the source’s velocity relative to the propagating medium

( $v_s$).

Answer to Problem 36P

The sign of the source’s velocity relative to the propagating medium ($v_s$) should be positive.

Explanation of Solution

Given Info:

$f_0=f_s\left(\frac{v+v_0}{v-v_s}\right)$

• $f_s$ is the frequency emitted by the source
• $f_0$ is the frequency detected by the observer
• $v$ is the speed of the wave in the propagating medium
• $v_0$ is the velocity of the observer relative to the medium
• $v_s$ is the source’s velocity relative to the propagating medium

Since, the observed frequency is increasing; the denominator of the above equation should be decreasing. Thus, the sign of the source’s velocity relative to the propagating medium ( $v_s$) should be positive.

Thus, the sign of the source’s velocity relative to the propagating medium ( $v_s$) should be positive.

Conclusion:

The sign of the source’s velocity relative to the propagating medium ( $v_s$) should be positive.

To determine

The frequency observed by the crewman on the red submarine.

Answer to Problem 36P

The frequency observed by the crewman on the red submarine is $5.30\times {10}^3\text{Hz}$.

Explanation of Solution

Given Info:

The speed of the yellow submarine is $11.0\text{m/s}$.

The frequency of the sonar is $5.27\times {10}^3\text{Hz}$.

The speed of the red submarine is $3.0\text{m/s}$ relative to the water in the same direction.

The speed of the sound in sea water is $1533\text{m/s}$.

Formula to calculate the frequency of the crewman is,

$f_0=f_s\left(\frac{v+v_0}{v-v_s}\right)$

• $f_s$ is the frequency emitted by the source
• $f_0$ is the frequency detected by the observer
• $v$ is the speed of the wave in the propagating medium
• $v_0$ is the velocity of the observer relative to the medium
• $v_s$ is the source’s velocity relative to the propagating medium

Substitute $5.27\times {10}^3\text{Hz}$ for $f_s$, $1533\text{m/s}$ for v, $-3.0\text{m/s}$ for $v_0$ and $11.0\text{m/s}$ for $v_s$ to find the frequency of the crew man,

$\begin{array}{c}{f}_0=\left(5.27\times {10}^3\text{Hz}\right)\left(\frac{1533\text{m/s}+\left(-3.0\text{m/s}\right)}{1533\text{m/s}-11.0\text{m/s}}\right)\\ =5.30\times {10}^3\text{Hz}\end{array}$

Thus, the frequency observed by the crewman on the red submarine is $5.30\times {10}^3\text{Hz}$

Conclusion:

The frequency observed by the crewman on the red submarine is $5.30\times {10}^3\text{Hz}$.