A source of sound is moving along a circular path of radius 3 m with an angular velocity of 10 rad/s

Q: Why can we assume the line of sight is horizontal?

Because the detector is located 'far away', the angle subtended by the source's circular path at the detector's position is extremely small, making the line of sight practically parallel to the horizontal axis.

Q: Why do the maximum and minimum frequencies occur at exactly $t = 3T/4$ and $t = T/4$?

Since the source and detector are perfectly synchronized with the same time period, their maximum horizontal speeds occur at the same instants. At $t = 3T/4$, both are moving towards each other, and at $t = T/4$, both are moving away.

Q: What would happen if the detector was not far away?

We would have to calculate the instantaneous angle of the line of sight and project both velocity vectors along it, which would make the frequency variation non-sinusoidal and much more complex.

Q: Why is the detector's maximum velocity $60\text{ m/s}$?

In simple harmonic motion, the maximum velocity is given by $v_{\max} = \omega A$. With $\omega = 10\text{ rad/s}$ and $A = 6\text{ m}$, we get $10 \times 6 = 60\text{ m/s}$.

Q: Does the speed of sound change because of the motion?

No, the speed of sound is a property of the medium (air) and remains constant at $340\text{ m/s}$ regardless of the motion of the source or detector.

Question

A source of sound is moving along a circular path of radius $3	ext{ m}$ with an angular velocity of $10	ext{ rad/s}$. A sound detector located far away from the source is executing linear simple harmonic motion along the line $BD$ with an amplitude $BC = CD = 6	ext{ m}$. The frequency of oscillation of the detector is $5/\pi$ per second. The source is at the point $A$ when the detector is at the point $B$. If the source emits a continuous sound wave of frequency $340	ext{ Hz}$, find the maximum and the minimum frequencies recorded by the detector. (Speed of sound = $340	ext{ m/s}$)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a vast, open field.

In front of you, a sound source is spinning in a perfect circle of radius $R = 3	ext{ m}$ at a rapid pace of $\omega_s = 10	ext{ rad/s}$.

Far away, aligned horizontally with this circle, a sound detector is sliding back and forth in simple harmonic motion (SHM) with an amplitude of $6	ext{ m}$.

This is not just a problem of simple motion; it is a beautifully synchronized cosmic dance.

Let's first look at the frequencies of these two motions.

The source rotates with an angular frequency of $\omega_s = 10	ext{ rad/s}$.

The detector oscillates with a frequency of $n_d = 5/\pi	ext{ Hz}$, which translates to an angular frequency of:

$$\omega_d = 2\pi n_d = 2\pi \left(\frac{5}{\pi}ight) = 10	ext{ rad/s}$$

They are perfectly synchronized! Both complete one full cycle in the exact same time period:

$$T = \frac{2\pi}{10} = \frac{\pi}{5}	ext{ s}$$

This synchronization means that their relative positions and velocities will repeat identically every cycle, allowing us to find the absolute maximum and minimum frequencies by looking at specific phases of their motion.

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### The Master Equations of Motion

Because the detector is located **far away**, we can make a brilliant geometric simplification.

The line of sight from the source to the detector is practically horizontal.

This means that any vertical component of the source's velocity does not contribute to the Doppler shift.

Only the horizontal components of their velocities along the line of sight matter.

Let's set up a coordinate system where the positive $x$-axis points to the right (towards the detector).

At $t = 0$, the source is at point $A$ (the rightmost point of the circle, $	heta = 0$) and is moving counter-clockwise.

Its horizontal velocity component at any time $t$ is:

$$v_{s,x}(t) = -v_s \sin(\omega t) = -30 \sin(10 t)	ext{ m/s}$$

At the same instant $t = 0$, the detector is at point $B$ (the leftmost extreme of its SHM).

Its displacement from the center $C$ is:

$$x(t) = -6 \cos(\omega t)	ext{ m}$$

Differentiating this displacement gives the velocity of the detector:

$$v_o(t) = \frac{dx}{dt} = 60 \sin(10 t)	ext{ m/s}$$

Notice how both velocities are beautifully coupled by the term $\sin(10 t)$.

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### Finding the Maximum Frequency

To get the absolute maximum frequency, we need the ultimate **double-approach** scenario.

The source must be moving towards the detector at its maximum speed, and the detector must be moving towards the source at its maximum speed.

This means we need $v_{s,x}(t)$ to be maximum positive ($+30	ext{ m/s}$) and $v_o(t)$ to be maximum negative ($-60	ext{ m/s}$).

Looking at our equations, this happens simultaneously when:

$$\sin(10 t) = -1$$

This occurs at:

$$10 t = \frac{3\pi}{2} \implies t = \frac{3T}{4}$$

At this precise moment, the source is at the bottom of the circle, moving directly to the right towards the detector.

Simultaneously, the detector is passing through the center $C$, rushing to the left towards the source.

Using the Doppler formula for mutual approach:

$$f_{\max} = f_0 \left(\frac{v + v_o}{v - v_s}ight)$$

Substituting the values:

$$f_{\max} = 340 \left(\frac{340 + 60}{340 - 30}ight) = 340 \left(\frac{400}{310}ight) \approx 438.7	ext{ Hz}$$

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### Finding the Minimum Frequency

For the absolute minimum frequency, we need the ultimate **double-recession** scenario.

Both the source and the detector must be moving away from each other at their maximum respective speeds.

This requires $v_{s,x}(t)$ to be maximum negative ($-30	ext{ m/s}$) and $v_o(t)$ to be maximum positive ($+60	ext{ m/s}$).

This occurs when:

$$\sin(10 t) = +1$$

Which happens at:

$$10 t = \frac{\pi}{2} \implies t = \frac{T}{4}$$

At this instant, the source is at the top of the circle, moving to the left away from the detector.

Meanwhile, the detector is at the center $C$, moving to the right away from the source.

Using the Doppler formula for mutual recession:

$$f_{\min} = f_0 \left(\frac{v - v_o}{v + v_s}ight)$$

Substituting the values:

$$f_{\min} = 340 \left(\frac{340 - 60}{340 + 30}ight) = 340 \left(\frac{280}{370}ight) \approx 257.3	ext{ Hz}$$

This completes our elegant journey through this synchronized acoustic system!

Visualized Solution

Understanding the Setup and Synchronized Motion

Calculating Velocities of Source and Detector

Analyzing Phase and Position at $t = 0$

Formulating Equations of Motion

Conditions for Maximum and Minimum Frequency

Finding the Time for Maximum Frequency

Calculating Maximum Frequency $f_{m a x}$

Finding the Time for Minimum Frequency

Calculating Minimum Frequency $f_{m i n}$

Exploring Variations and Extensions

The Sigma Insight: Doppler Effect

Analyzing the Setup

The Master Equations of Motion

Finding the Maximum Frequency

Finding the Minimum Frequency

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Comprehension Passage

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