A source (S) of sound has frequency 240 Hz. When the observer (O) and the source move towards each o

Q: Why is the sign in the denominator negative when the source approaches?

When the source moves towards the observer, it compresses the sound waves, effectively decreasing the wavelength. A smaller wavelength means a higher frequency. To make the overall fraction larger, the denominator must be smaller, hence the negative sign.

Q: Can we assume the speed of sound to be 340 m/s here?

No, you should never assume the speed of sound unless it's explicitly given. In this problem, the speed of sound ($c$) is not provided, but we can find its relationship with $v$ using the first case.

Q: What if the wind was blowing?

If wind is blowing with velocity $v_w$ from the source to the observer, the effective speed of sound becomes $c + v_w$. You would replace $c$ with $c + v_w$ in the formula. But here, the medium is stationary.

Q: Why do the $v$'s cancel out in the final step?

Because both the source and observer are moving at speeds proportional to the speed of sound (since $c = 11v$). The Doppler shift depends purely on the ratio of velocities, so the absolute value of $v$ isn't needed to find the final frequency.

Q: Is the Doppler effect applicable for light waves too?

Yes, but the formula is different because light doesn't require a medium and follows the theory of relativity. For sound, the velocities of the source and observer relative to the medium matter individually.

Question

A source (S) of sound has frequency 240 Hz. When the observer (O) and the source move towards each other at a speed v with respect to the ground (as shown in Case 1 in the figure), the observer measures the frequency of the sound to be 288 Hz. However, when the observer and the source move away from each other at the same speed v with respect to the ground (as shown in Case 2 in the figure), the observer measures the frequency of sound to be n Hz. The value of n is _____.

Sigma Percentile · Accepted Answer

### The Symphony of Relative Motion

Imagine standing on a quiet street when an ambulance races past you with its sirens blaring. You've likely noticed how the pitch of the siren sounds higher as it approaches and suddenly drops as it speeds away. This everyday phenomenon is the **Doppler Effect**, a beautiful interplay between the relative motion of a source and an observer.

In this problem, we are given a source emitting a sound of frequency $f_0 = 240 	ext{ Hz}$. We have two distinct scenarios: first, the source and observer are rushing towards each other, and second, they are retreating from one another. Our goal is to find the apparent frequency heard during their retreat.

### The Master Equation of Doppler Effect

To solve any Doppler effect problem, we rely on one master equation. The apparent frequency $f'$ heard by the observer is given by:

$$f' = f_0 \left( \frac{c \pm v_o}{c \mp v_s} ight)$$

Here, $c$ is the speed of sound in the medium, $v_o$ is the speed of the observer, and $v_s$ is the speed of the source. The true challenge lies not in the formula itself, but in the **sign convention**.

The rule of thumb is simple: any motion that tends to bring the source and observer closer together will increase the apparent frequency. Therefore, an approaching observer gets a positive sign in the numerator, and an approaching source gets a negative sign in the denominator.

### Case 1: The Approach

Let's analyze the first case. Both the source and the observer are moving towards each other with a speed $v$. Applying our sign convention, the observer's motion increases the frequency (so we use $+v$ in the numerator), and the source's motion also increases the frequency (so we use $-v$ in the denominator).

This gives us our first working equation:

$$f_1 = f_0 \left( \frac{c + v}{c - v} ight)$$

We are given that during this approach, the observer hears a frequency of $288 	ext{ Hz}$. Substituting the known values:

$$288 = 240 \left( \frac{c + v}{c - v} ight)$$

### Unlocking the Speed Ratio

This equation is a treasure map that leads us to the relationship between the speed of sound $c$ and the speed of the objects $v$. Let's simplify the fraction:

$$\frac{c + v}{c - v} = \frac{288}{240} = \frac{6}{5}$$

Now, we cross-multiply to solve for $c$ in terms of $v$:

$$5(c + v) = 6(c - v)$$

$$5c + 5v = 6c - 6v$$

Bringing the terms together, we find a remarkably clean relationship:

**$$c = 11v$$**

This tells us that the speed of sound is exactly 11 times the speed of the source and observer. We don't need the absolute values of these speeds; their ratio is all the power we need.

### Case 2: The Retreat

Now, the scene shifts. The source and observer have crossed paths and are now moving away from each other at the same speed $v$.

Because they are separating, the apparent frequency must drop. The observer moving away gets a negative sign in the numerator, and the source moving away gets a positive sign in the denominator. Our new equation is:

$$f_2 = f_0 \left( \frac{c - v}{c + v} ight)$$

### The Final Crescendo

We are ready for the final calculation. We substitute our golden ratio, $c = 11v$, into the new equation:

$$f_2 = 240 \left( \frac{11v - v}{11v + v} ight)$$

$$f_2 = 240 \left( \frac{10v}{12v} ight)$$

The variable $v$ elegantly cancels out, leaving us with a simple arithmetic step:

$$f_2 = 240 \left( \frac{5}{6} ight) = 200 	ext{ Hz}$$

Thus, when the source and observer are receding, the measured frequency drops to **$200 	ext{ Hz}$**. The value of $n$ is exactly 200. By carefully managing our sign conventions and leveraging the ratio of speeds, a seemingly complex relative motion problem unravels into a beautiful, logical conclusion.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e D o ppl er E f f ec t$

$T h e G e n er a l D o ppl er F or m u l a$

$C a se 1 : A pp r o a c hin g E a c h O t h er$

$S u b s t i t u t in g V a l u es f or C a se 1$

$F in d in g t h e R e l a t i o n b e tw ee n c an d v$

$C a se 2 : R ece d in g f r o m E a c h O t h er$

$C a l c u l a t in g t h e F ina l F r e q u e n cy$

$C o n c l u s i o n$

The Sigma Insight: Doppler Effect

The Symphony of Relative Motion

The Master Equation of Doppler Effect

Case 1

Unlocking the Speed Ratio

Case 2

The Final Crescendo

Similar Questions

A stationary source emits sound waves of frequency $500 Hz$ . Two observers moving along a line passing through the source detect sound to be of frequencies $480 Hz$ and $530 Hz$ . Their respective speeds are in $ms^{- 1}$ , (Take, speed of sound $= 300 m/s$ )

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 $330 m/s$ ?

An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

A train is moving on a straight track with speed $20 ms^{- 1}$ . It is blowing its whistle at the frequency of $1000 Hz$ . The percentage change in the frequency heard by a person standing near the track as the train passes him is close to (speed of sound = $320 ms^{- 1}$ )