A police car moving at 22 m/s chases a motorcyclist. The police man sounds his horn at 176 Hz, while

Q: Why did we use a minus sign in both the numerator and denominator for the police car's frequency?

The sign convention in the Doppler formula is determined by the direction of motion. The police car (source) is moving towards the motorcyclist, which increases frequency, requiring a minus sign in the denominator. The motorcyclist (observer) is moving away from the police car, which decreases frequency, requiring a minus sign in the numerator.

Q: What would happen if the motorcyclist was moving away from the stationary siren instead of towards it?

If the motorcyclist were moving away from the stationary siren, the apparent frequency of the siren would decrease, and we would use a minus sign in the numerator: $f_2' = f_2 \left(\frac{v_{\text{sound}} - v}{v_{\text{sound}}}\right)$.

Q: How does the beat frequency relate to the two individual frequencies?

The beat frequency is the absolute difference between the two frequencies heard by the observer: $f_{\text{beat}} = |f_1' - f_2'|$. For zero beats, this difference must be zero, meaning $f_1' = f_2'$.

Q: Is the speed of sound always constant in these problems?

Yes, unless specified otherwise (such as temperature changes or wind blowing), the speed of sound is assumed to be constant throughout the medium. Here, it is given as $330\text{ m/s}$.

Q: Can we solve this problem using relative velocity directly?

No, because the Doppler effect depends on the velocities of the source and observer relative to the medium (air), not just relative to each other. This is why we must use the absolute velocities of the source and observer in the Doppler formula.

Question

A police car moving at $22	ext{ m/s}$ chases a motorcyclist. The police man sounds his horn at $176	ext{ Hz}$, while both of them move towards a stationary siren of frequency $165	ext{ Hz}$. Calculate the speed of the motorcycle, if it is given that the motorcyclist does not observe any beats (speed of sound = $330	ext{ m/s}$).

Options:

(A) $33	ext{ m/s}$

(B) $22	ext{ m/s}$

(C) Zero

(D) $11	ext{ m/s}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on a long, straight highway.

To your left, a police car is speeding forward at a constant velocity of $22	ext{ m/s}$, sounding its horn at a frequency of $176	ext{ Hz}$.

In the middle, a motorcyclist is riding in the same direction with an unknown speed $v$.

To the far right, a stationary siren is blaring at a frequency of $165	ext{ Hz}$.

Both the police car and the motorcyclist are moving towards this stationary siren.

Our goal is to find the speed $v$ of the motorcycle such that the rider hears absolutely **no beats**.

### The Physics of Beats and Doppler Shift

What does it mean physically when we say "no beats are heard"?

Beats are a wave interference phenomenon.

When two sound waves of slightly different frequencies reach an observer, they superimpose to produce a sound that periodically rises and falls in intensity.

The frequency of this beating is simply the difference between the two frequencies:

$$f_{	ext{beat}} = |f_1' - f_2'|$$

If the motorcyclist hears **no beats**, the beat frequency must be exactly zero.

This implies that the two apparent frequencies heard by the motorcyclist must be perfectly identical:

$$f_1' = f_2'$$

Here, $f_1'$ is the apparent frequency of the police car's horn as heard by the motorcyclist, and $f_2'$ is the apparent frequency of the stationary siren as heard by the motorcyclist.

To find these apparent frequencies, we must apply the **Doppler Effect** formula:

$$f' = f_0 \left(\frac{v_{	ext{sound}} \pm v_o}{v_{	ext{sound}} \pm v_s}ight)$$

Let's carefully analyze the sign conventions for both sources.

### Apparent Frequency of the Police Horn ($f_1'$)

Let's look at the interaction between the police car (source) and the motorcyclist (observer):

1. The police car is chasing the motorcyclist. Since the source is moving *towards* the observer, this motion tends to increase the frequency. Therefore, we use a minus sign in the denominator: $v_{	ext{sound}} - v_p$.

2. The motorcyclist is moving *away* from the police car. Since the observer is moving away from the source, this motion tends to decrease the frequency. Therefore, we use a minus sign in the numerator: $v_{	ext{sound}} - v$.

Combining these, we get:

$$f_1' = f_1 \left(\frac{v_{	ext{sound}} - v}{v_{	ext{sound}} - v_p}ight)$$

Substituting the given values ($f_1 = 176	ext{ Hz}$, $v_{	ext{sound}} = 330	ext{ m/s}$, and $v_p = 22	ext{ m/s}$):

$$f_1' = 176 \left(\frac{330 - v}{330 - 22}ight)$$

$$f_1' = 176 \left(\frac{330 - v}{308}ight)$$

Notice that $176$ and $308$ share a common factor of $44$:

$$\frac{176}{44} = 4, \quad \frac{308}{44} = 7$$

Thus, the expression simplifies beautifully to:

$$f_1' = \frac{4}{7}(330 - v)$$

### Apparent Frequency of the Stationary Siren ($f_2'$)

Now, let's look at the interaction between the stationary siren (source) and the motorcyclist (observer):

1. The siren is stationary, so $v_s = 0$.

2. The motorcyclist is moving *towards* the stationary siren. Since the observer is moving towards the source, this motion tends to increase the frequency. Therefore, we use a plus sign in the numerator: $v_{	ext{sound}} + v$.

Combining these, we get:

$$f_2' = f_2 \left(\frac{v_{	ext{sound}} + v}{v_{	ext{sound}}}ight)$$

Substituting the given values ($f_2 = 165	ext{ Hz}$ and $v_{	ext{sound}} = 330	ext{ m/s}$):

$$f_2' = 165 \left(\frac{330 + v}{330}ight)$$

Since $165$ is exactly half of $330$, this simplifies directly to:

$$f_2' = \frac{1}{2}(330 + v)$$

### Solving for the Motorcycle's Speed

Since the motorcyclist hears no beats, we equate the two simplified apparent frequencies:

$$\frac{4}{7}(330 - v) = \frac{1}{2}(330 + v)$$

To clear the fractions, we cross-multiply by multiplying both sides by $14$:

$$8(330 - v) = 7(330 + v)$$

Expanding both sides:

$$8 	imes 330 - 8v = 7 	imes 330 + 7v$$

Rearranging the terms to group the unknown variable $v$ on one side:

$$8 	imes 330 - 7 	imes 330 = 7v + 8v$$

$$(8 - 7) 	imes 330 = 15v$$

$$330 = 15v$$

$$v = \frac{330}{15} = 22	ext{ m/s}$$

Thus, the speed of the motorcycle must be exactly **$22	ext{ m/s}$**.

This corresponds perfectly to **Option (b)**.

Select Answer:

Visualized Solution

Visualizing the Doppler Setup

The Doppler Effect Formula

The Condition for No Beats

Apparent Frequency of the Police Horn

Substituting Values for $f_{1}^{'}$

Simplifying the Police Horn Expression

Apparent Frequency of the Stationary Siren

Substituting Values for $f_{2}^{'}$

Simplifying the Siren Expression

Equating the Two Frequencies

Cross-Multiplying to Clear Fractions

Expanding and Grouping Terms

Solving for the Speed of the Motorcycle

Final Answer Verification

Exploring Further Scenarios

The Sigma Insight: Doppler Effect

Analyzing the Setup

The Physics of Beats and Doppler Shift

Apparent Frequency of the Police Horn ( $f_{1}^{'}$ )

Apparent Frequency of the Stationary Siren ( $f_{2}^{'}$ )

Solving for the Motorcycle's Speed

Similar Questions

A police car with a siren of frequency $8 kHz$ is moving with uniform velocity $36 km/h$ towards a tall building which reflects the sound waves. The speed of sound in air is $320 m/s$ . The frequency of the siren heard by the car driver is

Two cars are approaching each other at an equal speed of $7.2 km/h$ . When they see each other, both blow horns having frequency of $676 Hz$ . The beat frequency heard by each driver will be ......... $Hz$ . [Velocity of sound in air is $340 m/s$ .]

A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from $440 Hz$ to $480 Hz$ , when it gets reflected from the wall. If the speed of sound in air is $345 m/s$ , then the speed of the car is

The frequency of a car horn encountered a change from 400 Hz to 500 Hz, when the car approaches a vertical wall. If the speed of sound is 330 m/s, then the speed of car is …… km/h.

The driver of a bus approaching a big wall notices that the frequency of his bus's horn changes from $420 Hz$ to $490 Hz$ , when he hears it after it gets reflected from the wall. Find the speed of the bus if speed of the sound is $330 ms^{- 1}$ .

Two sources of sound $S_{1}$ and $S_{2}$ produce sound waves of same frequency $660 Hz$ . A listener is moving from source $S_{1}$ towards $S_{2}$ with a constant speed $u m/s$ and he hears $10 beats/s$ . The velocity of sound is $330 m/s$ . Then, $u$ equal to

A source of sound of frequency $256 Hz$ is moving rapidly towards a wall with a velocity of $5 m/s$ . How many beats per second will be heard by the observer on source itself if sound travels at a speed of $330 m/s$ ?