Doppler Effect Problems with Solutions for College Physics
Here, you can solve some problems on the Doppler effect in college physics. The solutions are presented in a way that is comprehensive and self-taught.
When there is a relative motion between a sound (or other wave types) source and observer, the pitch (frequency) the observer hears varies by \[f_O=\frac{v\pm v_O}{v\pm v_S} f_S\] where $\rm O$ and $\rm S$ stand for observer and source, and $v$ is the speed of sound in still air.
In all Doppler effect questions, the most important thing about the formula above is to recognize when a positive or negative sign must be used for the velocities of the source and the observer.
Always draw an arrow pointing from the source toward the observer (receiver) and take this direction as a reference. If the observer or source velocities are in this direction, then we must select the negative sign in the above formula for both, otherwise the positive is substituted.
In the following, practice some practice problems to master this topic.
Doppler Effect Problems
Problem (1): The siren of a police car moving at a constant speed of $30\,\rm m/s$ emits a wave with a frequency of $1200\,\rm Hz$. What frequency will an individual at rest hear when the police car moves (a) toward her, or (b) away from her? The speed of sound is $340\,\rm m/s$, and the air is still.
Solution: (a) Consider the following figure in which a siren police car approaches a person at rest from the left.
Draw a vector from the siren to the person. This is the reference direction for our problem. The police car's velocity is in this direction, so we must choose the positive sign for the velocity of the police car. On the other hand, the person (or observer) is at rest, so $v_O=0$. Substituting these into the following formula gives, \begin{align*} f_O &=\frac{v\pm v_O}{v\pm v_S} f_S \\\\ &=\frac{v}{v-v_S}f_S \\\\ &=\frac{340}{340-30}(1200) \\\\ &\approx 1316\quad \rm Hz \end{align*} Thus, if a siren emitting $1200\,\rm Hz$ approaches (with $30\,\rm m/s$ toward a person at rest, he/she hears a sound of a frequency of about $1316\,\rm Hz$.
(b) Now, the siren is moving away from the observer, say toward the right.
Draw a vector from right to left and compare the sign of all other velocities with this direction. In this case, the siren's velocity is in the opposite direction of this vector, so we must choose a positive for its speed. Before solving, we expect a lower pitch (frequency) is perceived by the person. \begin{align*} f_O &=\frac{v\pm v_O}{v\pm v_S} f_S \\\\ &=\frac{v}{v+v_S}f_S \\\\ &=\frac{340}{340+30}(1200) \\\\ &\approx 1102\quad \rm Hz \end{align*} As expected.
Problem (2): A sound wave of frequency $525\,\rm Hz$ is emitted by a stationary source toward an observer who is moving away with a constant speed of $25\,\rm m/s$.
(a) What is the frequency perceived by the observer?
(b) Calculate the wavelength of the sound as measured by the source and the observer.
Solution: The given information is: the source frequency $f_S=525\,\rm Hz$, the source's speed $v_S=0$, the observer's speed $v_O=25\,\rm m/s$, and the sound speed in still air $v=340\,\rm m/s$.
(a) Assume the observer is moving toward the right. The vector originating from the source to the observer is to the right, in the same direction as the observer's velocity. Hence, a minus sign should be included for $v_O$ in the Doppler effect formula. \begin{align*} f_O&=\frac{v\pm v_O}{v\pm v_S} f_S \\\\ &=\frac{v-v_O}{v}f_S \\\\ &=\frac{340-25}{340} 525 \\\\ &\approx 486 \quad \rm Hz \end{align*} As expected, since the observer is moving away from a source at rest, it detects a lower frequency than the original one emitted by the source.
(b) The three main characteristics of a wave—wavelength, frequency, and wave speed are related together by \[\lambda=\frac{v}{f}\] According to this equation, the wavelength that the source at rest measures for its emitting sound wave is \[\lambda_S=\frac{v}{f_S}=\frac{340}{525}=0.65\,\rm m\] which equals to nearly $\lambda_S=65\,\rm cm$.
On the other hand, the wavelength that is measured by the moving observer is also found to be \[\lambda_S=\frac{v}{f_S}=\frac{340}{486}=0.70\,\rm m\] That equals to about $\lambda_O=70\,\rm cm$.
Problem (3): Suppose a police car is at rest and emits at $1200\,\rm Hz$. What frequency would a person hear if he were moving at $30\,\rm m/s$ (a) toward it or (b) away from it?
Solution: This problem is exactly the opposite of the previous problem. Here, the siren (emitter) is at rest; while the person (receiver) is moving.
(a) Suppose the person is moving toward the siren from the left side, as shown in the figure below.
Draw a vector from the siren to the person (in this case, in the $-x$ direction). The siren is at rest, so $v_S=0$. The person's velocity is in the opposite direction of the above reference vector, so it must be included with a positive in the Doppler effect formula. \begin{align*} f_O &=\frac{v\pm v_O}{v\pm v_S} f_S \\\\ &=\frac{v+v_O}{v}f_S \\\\ &=\frac{340+30}{340}(1200) \\\\ &\approx 1305\quad \rm Hz \end{align*} An increase in the original frequency emitted by the siren is perceived by an approaching person, as expected.
(b) Now, consider the case that the person is moving away from the siren at rest and is on the right side of it.
The vector pointing from the siren to the moving person, this time, is toward the right, in the same direction as the person is traveling. Thus, the person's velocity, $v_O$, must be included with a negative in the formula. \begin{align*} f_O &=\frac{v\pm v_O}{v\pm v_S} f_S \\\\ &=\frac{v-v_O}{v}f_S \\\\ &=\frac{340-30}{340}(1200) \\\\ &\approx 1094\quad \rm Hz \end{align*} As expected for a moving receiver (person) away from a static emitting source.
Problem (4): The siren of a police car moving away from a person at a speed of $35\,\rm m/s$ emits at $550\,\rm Hz$. The person is also moving toward the siren at a speed of $5\,\rm m/s$. Assume the speed of sound in the air to be $340\,\rm m/s$. What frequency would the person hear?
Solution: In this Doppler effect problem, both the source and listener (receiver) are moving. Assume the siren is moving to the right. The vector pointing from the source (siren) to the listener (receiver) is toward the left. We call this the reference vector.
The siren and the listener (observer) are traveling in the opposite direction of the reference vector, so we choose positive signs for them.
Taking into account all these considerations in the Doppler effect formula, we have \begin{align*} f_O &=\frac{v\pm v_O}{v\pm v_S} f_S \\\\ &=\frac{v+v_O}{v+v_S}f_S \\\\ &=\frac{340+5}{340+35}(550) \\\\ &=506\quad \rm Hz \end{align*} Let's check our numerical result. Both the siren and the person are moving to the right, but the person moves at a slower speed than the car, so the siren is gradually moving away from the person. In this case, we expect that the person will hear a reduction in the original frequency emitted from the siren as we calculated.
Problem (5): Two cars are moving (a) toward each other, (b) away from each other. The car $1$ has a speed of $120\,\rm km/h$ while the car $2$ moves at $95\,\rm km/h$. Assume car $1$ blows its horn and emits a sound at a frequency of $400\,\rm Hz$. What frequency is heard by the driver of the car $2$?
Solution: Both cars are moving at different velocities in $\rm km/h$. First, convert them into the SI units of velocity to be able to use them in the Doppler effect formula. \begin{gather*} v_1=120\times \frac{1000}{3600}\approx 33 \,\rm m/s \\\\ v_2=95\times \frac{1000}{3600}\approx 26 \,\rm m/s \end{gather*}
(a) Both cars are moving toward each other, as shown in the figure below. In this problem, car $1$ is assumed to be the source, and the other is the observer (receiver). The vector extending from the source to the receiver is to the right. Cars $1$ and $2$ move in the same and opposite directions of the reference vector, respectively. Hence, they must be included with a negative and a positive in the formula, respectively.
Substituting the numerical values into the known formula gives the frequency heard by the driver of car $2$ (receiver) as \begin{align*} f_O &=\frac{v\pm v_O}{v\pm v_S} f_S \\\\ &=\frac{v+v_O}{v-v_S}f_S \\\\ &=\frac{340+26}{340-33}(400) \\\\ &\approx 477 \quad \rm Hz \end{align*} As expected.
(b) In this case, the two cars are moving away from each other, as depicted in the figure below.
Again, the reference vector extends from the car $1$ (source) to the car $2$ (observer), which is toward the right. The car $1$ (source) moves in the opposite direction, but the car $2$ moves in the same direction as this reference vector. Thus, the speed of car $1$ (source) is entered with a positive in the formula, and the speed of car $2$ (observer) is accompanied by a negative. \begin{align*} f_O &=\frac{v\pm v_O}{v\pm v_S} f_S \\\\ &=\frac{v-v_O}{v+v_S}f_S \\\\ &=\frac{340-26}{340+33}(400) \\\\ &\approx 335 \quad \rm Hz \end{align*}
Problem (6): Suppose you are standing by a railway. You record the train whistle at $554\,\rm Hz$ when the train is approaching you, and $356\,\rm Hz$ when the train is moving away from you. How fast does the train move? (Take the speed of sound in air to be $v=340\,\rm m/s$).
Solution: This is a slightly more difficult Doppler effect problem since you should apply twice the Doppler effect formula. The problem is depicted in the following figure. The person (receiver or listener) is at rest, $v_O=0$, and the train is approaching and receding from the person at different frequencies.
Let $f_{O-I}$ and $f_{O-II}$ be the recorded frequencies by the person when the train is approaching and receding, respectively. In the approaching case, the vector pointing from the source to the observer is toward the right (the red arrow), in the same direction as the source velocity. Thus, the frequency perceived by the person is calculated by \begin{align*} f_{O-I} &=\frac{v\pm v_O}{v\pm v_S} f_S \\\\ &=\frac{v}{v-v_S}f_S \end{align*} where $f_S$ is the original frequency emitted by the train whistle (source).
In the receding case, the vector pointing from the source to the observer is to the left, in the opposite direction of the source velocity. Thus, $v_S$ should be chosen by a positive. \begin{align*} f_{O-II} &=\frac{v\pm v_O}{v\pm v_S} f_S \\\\ &=\frac{v}{v+v_S}f_S \end{align*} Dividing these two expressions by each other and canceling the common factors, gives us a relation in which the speed of the train is unknown. \[\frac{f_{O-I}}{f_{O-II}}=\frac{v+v_S}{v-v_S}\] Substituting the numerical values into this and solving for $v_S$ gives \begin{gather*} \frac{554}{356}=\frac{340+v_S}{340-v_S} \\\\ 554(340-v_S)=356(340+v_S) \\\\ 340(554-356)=v_S(356+554) \\\\ \Rightarrow \boxed{v_S\approx 74\,\rm m/s} \end{gather*} Hence, the train approaches the person at a constant speed of $74\,\rm m/s$ and then moves away from him.
Problem (7): Two car horns each have a frequency of $375\,\rm Hz$. A stationary person is between them and is moving toward the right at a speed of $10\,\rm m/s$. Suppose the car $A$ is at rest while the car $2$ is moving toward the right (and away from the observer) with a speed of $25\,\rm m/s$. (a) What is the frequency of $A$ heard by the person? (b) What frequency is heard from $B$ by the person? (c) What is the beat frequency counted by the observer?
Solution: The whole problem is illustrated in the following figure.
First of all, on both sides, draw vectors (called reference vectors) pointing from the source to the observer (person). On the left side, this vector points to the right, and on the right side, it points toward the left.
The frequency detected by the person from the car $A$ (source) is calculated as below \begin{align*} f_{S_A} &=\frac{v\pm v_O}{v\pm v_{S_A}} f_{S_A} \\\\ &=\frac{v-v_O}{v}f_{S_A} \\\\ &=\frac{340-10}{340}(375) \\\\&\approx 364\,\rm Hz \end{align*} Because the reference vector and the person's velocity are in the same direction, a minus sign should be chosen for $v_O$ in this part of the problem.
(b) In this section, both the person (receiver or listener) and the source (car's horn) are moving in the same direction. On the other hand, the reference vector, here, points to the left, in the opposite direction of the source and the person, so a positive should be chosen for the velocities of the person (listener) and the source. \begin{align*} f_{O_B} &=\frac{v\pm v_O}{v\pm v_{S_B}} f_{S_B} \\\\ &=\frac{v+v_O}{v-v_{S_B}}f_{S_B} \\\\ &=\frac{340+10}{340+25}(375) \\\\&\approx 360\,\rm Hz \end{align*} As expected, because the velocity of the car $B$ is greater than that of the person. The source is gradually moving away from the observer, hence, a lower pitch (frequency) is heard.
(c) This part is related to a beat frequency problem. Recall that the interference (or superposition) of two waves with slightly different frequencies gives a new wave whose loudness varies in time with a frequency of $f_{beat}=|f_2-f_1|$, which is called beat frequency. Note that the absolute value of the difference between frequencies matters for finding the beat frequency. Substituting the numerical values into that formula gives \[f_{beat}=|360-364|=4\,\rm Hz\]
Problem (8): Suppose you hold a tuning fork vibrating at $245\,\rm Hz$ and are moving toward a wall with a constant speed of $6\,\rm m/s$. What frequency do you hear when the wave is reflected backward?
Solution: This problem is made up of two parts. Part $1$ is when you are approaching the wall. In this case, you are the source (emitter), and the wall is the receiver (observer). The vector pointing from the source to the observer is toward the right as shown in the figure below.
The source's velocity and this vector are in the same direction, so $v_S$ must be chosen to be negative. \begin{align*} f_O &=\frac{v\pm v_O}{v\pm v_S} f_S \\\\ &=\frac{v}{v-v_S}f_S \\\\ &=\frac{v}{v-6}f_S \end{align*} where $f_S$ is the frequency at which the tuning fork vibrates and $f_O$ is the frequency received by the wall.
Part $2$ is formed when the sound wave is reflected back from the wall (which is called an echo). In this case, the wall acts as an emitter (source) of a wave with a frequency of $f_{S'}$ and the original source (you in the first part) will act as the new observer (receiver). This new observer (you) perceives a frequency of $f_{O'}$. Now draw a vector from this new source to the new observer that points to the left. The new observer is moving in the opposite direction of this vector, so it should be accompanied by a positive. \begin{align*} f_{O'} &=\frac{v\pm v_{O'}}{v\pm v_{S'}} f_{S'} \\\\ &=\frac{v+v_{O'}}{v}f_{S'} \\\\ &=\frac{v+6}{v}f_{S'} \end{align*} where $f_{S'}$ is the frequency of the wave which is reflected back from the wall.
It is clear that the incoming frequency $f_O$ (in part $1$) and the reflected back frequency $f_{S'}$ (in part $2$) are the same; $f_O=f_{S'}$. The frequency of a wave does not change upon striking a barrier, so the incoming and reflected waves have the same frequencies. By combining the two above expressions, we reach the following final expression \begin{gather*} f_{O'}=\frac{v+v_{O'}}{v}\times \frac{v}{v-v_S} f_S \\\\ \Rightarrow \boxed{f_{O'}=\frac{v+v_{O'}}{v-v_S} f_s} \end{gather*} Substituting the numerical values and solving for the echo frequency heard by the person $f_{O'}$, we will have \begin{align*} f_{O'}&=\frac{v+v_{O'}}{v-v_S} f_s \\\\ &=\frac{340+6}{340-6}(245) \\\\ &\approx 254\quad \rm Hz \end{align*}
Problem (9): Suppose the horn frequency of two cars is the same. One is at rest, and the other is traveling toward the first at a constant speed of $25\,\rm m/s$. In this situation, the driver at rest detects a beat frequency of $3.5\,\rm Hz$. What frequency do the horns emit?
Solution: The difference between two nearly equal frequencies is called the beat frequency, $f_{beat}=|f_1-f_2|$ where $|\cdots|$ gives the absolute value of this subtraction. The driver at rest detects its own original horn frequency but hears a changed frequency for the other car. Our task is to find this changed frequency detected by the driver at rest.
The stationary driver is considered to be an observer ($v_O=0$) and hears a frequency of $f_O$. The other driver, then, is an emitter having a velocity of $v_S=25\,\rm m/s$ and emitting a sound of frequency $f_S$. According to the Doppler effect, $f_O$ is calculated by \[f_O =\frac{v\pm v_O}{v\pm v_S} f_S =\frac{v}{v-v_S} \] This is the frequency that the driver at rest hears from the other car. The difference between this frequency and the own horn frequency gives the beat frequency. \begin{align*} f_{beat}&=|f_O-f_S| \\\\ &=\left|\frac{v}{v-v_S}f_S-f_S \right| \\\\ &=f_S\left|\frac{v}{v-v_S}-1\right| \\\\ &=f_S \left|\frac{v_S}{v-v_S}\right| \end{align*} Substituting the numerical values into it and solving for the original horn frequency of these two cars yields \begin{gather*} f_{beat}=\frac{v_S}{v-v_S} f_S \\\\ 3.5=\frac{25}{340-25} f_S \\\\ \rightarrow (3.5)(340-25)=25f_S \\\\ \Rightarrow \boxed{f_S=44.1\,\rm Hz} \end{gather*} where in the above, we omitted $|\cdots|$ since the result is positive.
Problem (10): Suppose the frequency of a car's horn at rest is $350\,\rm Hz$. The car's driver sounds this horn, while it is moving. A bicyclist moving in the same direction as the car with one-fourth the car's speed hears a frequency of $325\,\rm Hz$. (a) Is the bike moving in front of or behind the car? (b) What is the speed of the car?
Solution: (a) To determine whether the bike is ahead of or behind the car, let's explore each case separately.
Assume the bike is moving at one-fourth the car's speed ahead of the car. Because the car is faster than the bike, the car gradually reaches the bike. In this case, we expect that the bicyclist will hear an increasing frequency, much like in the situation where an ambulance is approaching. Hence, this is not the case since the frequency heard by the cyclist is lower than that of the car.
Therefore, the other option of the bike moving behind the car is the correct result. Because the bike moves slower than the car, it is actually always getting further away from the car and subsequently hears a lower pitch (frequency).
(b) The actual situation of this problem is depicted in the following figure.
The car is the source, and the bike is an observer. The vector pointing from the source to the observer gives a hint to determining which sign should be included in the Doppler effect formula. This vector, in this case, is toward the left. Both the car and the bike move in the opposite direction of this vector, so a positive should be included for $v_S$ and $v_O$. \begin{gather*} f_O =\frac{v\pm v_O}{v\pm v_S} f_S \\\\ f_O=\frac{v+v_O}{v+v_S} f_S \end{gather*} Assume the car's speed to be $v_S=v$, so the bike's speed is $v_O=\frac 14 v$. Substituting the given numerical values into the above formula and solving for $v$, we will have \begin{gather*} 325=\frac{340+\frac 14 v}{340+v} 350 \\\\ \Rightarrow \boxed{v\approx 36\,\rm m/s} \end{gather*}
Summary:
In this article, we learned how to solve the Doppler effect problems step-by-step with detailed solutions. The most important step in solving the Doppler effect questions is to detect the source and the observer, then draw a vector from the source to the observer.
The above vector shows the direction of the sound wave toward the observer. If all velocities (source $v_S$ and observer $v_O$) involved in the Doppler effect formula are in the same direction as the reference vector, then a minus sign should be inserted and vice versa. \[f_O=\frac{v\pm v_O}{v\pm v_S}f_S\] where $v$ is the speed of the sound wave. $f_O$ and $f_S$ are the frequencies heard by an observer and the frequency emitted by a source, respectively.
Author: Dr. Ali Nemati
Published: 4/9/2022
