A whistle emitting a sound of frequency 440 Hz is tied to a string of 1.5 m length and rotated with

Q: Why do the maximum and minimum frequencies occur exactly at the tangent points?

Because the observer is at a very large distance, the lines of sight to all points on the circle are nearly parallel. Thus, the component of velocity along the line of sight is maximum (equal to the full speed $v_s$) at the points where the motion is purely horizontal (towards or away from the observer).

Q: What if the observer is not at a very large distance?

If the observer is close, the line of sight changes angle continuously. We would have to use the component of velocity along the line of sight, $v_s \cos\theta$, where $\theta$ is the angle between the velocity vector and the line of sight at any instant.

Q: Does the frequency change abruptly between the maximum and minimum values?

No, the frequency changes continuously and smoothly as a sinusoidal-like function of time because the component of velocity along the line of sight varies continuously as the source rotates.

Q: Why is the speed of sound in the numerator not affected by the source's motion?

The speed of sound in a medium depends only on the properties of the medium (like temperature and pressure) and is entirely independent of the motion of the source or the observer.

Q: What happens if the whistle rotates faster than the speed of sound?

If $v_s > v$, a sonic boom (shock wave) is created. The Doppler formula in its standard form breaks down because the source outruns its own sound waves, creating a cone of constructive interference.

Question

A whistle emitting a sound of frequency $440	ext{ Hz}$ is tied to a string of $1.5	ext{ m}$ length and rotated with an angular velocity of $20	ext{ rad/s}$ in the horizontal plane. Calculate the range of frequencies heard by an observer stationed at a large distance from the whistle. (Speed of sound = $330	ext{ m/s}$).

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a whistle tied to a string, spinning in a horizontal circle. This is a classic physics problem that beautifully marries **uniform circular motion** with the **Doppler effect** of sound.

Let's break down the parameters given to us:
- The natural frequency of the whistle is $f = 440	ext{ Hz}$.
- The length of the string, which acts as the radius of the circular path, is $R = 1.5	ext{ m}$.
- The angular velocity of rotation is $\omega = 20	ext{ rad/s}$.
- The speed of sound in air is $v = 330	ext{ m/s}$.

We are asked to find the range of frequencies heard by an observer stationed at a **large distance** from the whistle.

This "large distance" assumption is a crucial simplification. It means that the line of sight from the observer to any point on the circular path is practically parallel. Therefore, the maximum and minimum frequency shifts will occur exactly when the whistle is moving directly towards or directly away from the observer.

### Finding the Speed of the Source

Before we can apply the Doppler effect equations, we need to know how fast the whistle is actually moving along its circular path.

In uniform circular motion, the linear speed $v_s$ of an object is related to its angular velocity $\omega$ and radius $R$ by the simple relation:

$$v_s = R\omega$$

Substituting the given values into this equation:

$$v_s = 1.5	ext{ m} 	imes 20	ext{ rad/s} = 30	ext{ m/s}$$

So, the whistle is traveling at a constant speed of $30	ext{ m/s}$ along the circumference of the circle.

### The Doppler Effect Equations

The general formula for the Doppler-shifted frequency $f'$ heard by a stationary observer when the source is moving is:

$$f' = f \left(\frac{v}{v \mp v_s}ight)$$

Where:
- $v$ is the speed of sound ($330	ext{ m/s}$).
- $v_s$ is the speed of the source ($30	ext{ m/s}$).
- The negative sign in the denominator is used when the source is moving **towards** the observer (resulting in a higher frequency).
- The positive sign in the denominator is used when the source is moving **away** from the observer (resulting in a lower frequency).

### Calculating the Maximum Frequency

The maximum frequency $f_{	ext{max}}$ is perceived at the instant when the whistle is at the top of its circular path (point $P$), where its instantaneous velocity vector points directly towards the observer.

Using the subtraction sign in the denominator:

$$f_{	ext{max}} = f \left(\frac{v}{v - v_s}ight)$$

Substituting our values:

$$f_{	ext{max}} = 440 \left(\frac{330}{330 - 30}ight)$$

$$f_{	ext{max}} = 440 \left(\frac{330}{300}ight) = 440 	imes 1.1 = 484	ext{ Hz}$$

### Calculating the Minimum Frequency

Conversely, the minimum frequency $f_{	ext{min}}$ is heard when the whistle is at the bottom of its circular path (point $Q$), where its instantaneous velocity vector points directly away from the observer.

Using the addition sign in the denominator:

$$f_{	ext{min}} = f \left(\frac{v}{v + v_s}ight)$$

Substituting our values:

$$f_{	ext{min}} = 440 \left(\frac{330}{330 + 30}ight)$$

$$f_{	ext{min}} = 440 \left(\frac{330}{360}ight) = 440 	imes 0.9167 \approx 403.33	ext{ Hz}$$

### Conclusion

As the whistle continuously rotates, the frequency heard by the distant observer fluctuates smoothly and periodically between these two extreme limits.

Therefore, the range of frequencies heard by the observer is from **$403.3	ext{ Hz}$ to $484	ext{ Hz}$**.

Visualized Solution

Visualizing the Circular Motion

The Doppler Effect Principle

Relating Angular and Linear Velocity

Calculating the Source Speed

Condition for Maximum Frequency

Calculating Maximum Frequency

Condition for Minimum Frequency

Calculating Minimum Frequency

The Perceived Frequency Range

The Way Forward

The Sigma Insight: Doppler Effect

Analyzing the Setup

Finding the Speed of the Source

The Doppler Effect Equations

Calculating the Maximum Frequency

Calculating the Minimum Frequency

Conclusion

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