A stationary observer receives sound from two identical tuning forks, one of which approaches and th

Q: Why do we subtract the frequencies to get the beat frequency?

When two sound waves with slightly different frequencies reach your ear, they constructively and destructively interfere. This creates an envelope that pulses at a rate equal to the difference between the two frequencies, which we hear as beats.

Q: Why did we ignore $v_s^2$ in the denominator?

The problem states $v_s \ll v$. When you square a very small number, it becomes even smaller. For example, if $v_s$ is $1$ and $v$ is $350$, $v_s^2$ is $1$ and $v^2$ is $122500$. Subtracting $1$ from $122500$ makes virtually no difference, so we approximate $v^2 - v_s^2 \approx v^2$ to make the math much easier.

Q: What if the tuning forks were moving towards the observer at the speed of sound?

If $v_s = v$, the denominator for the approaching fork ($v - v_s$) becomes zero, meaning the apparent frequency approaches infinity. Physically, this represents a sonic boom or shock wave, where all the sound waves pile up on top of each other.

Q: Does the initial distance of the tuning forks from the observer matter?

No, the Doppler shift depends entirely on the relative velocity between the source and the observer, not their separation distance. As long as the velocity is constant, the apparent frequency remains constant.

Q: Can we use the binomial approximation $(1-x)^{-1} \approx 1+x$ here?

Yes, absolutely! You can write $\frac{v}{v-v_s} = (1 - \frac{v_s}{v})^{-1} \approx 1 + \frac{v_s}{v}$. Doing this for both frequencies and subtracting them will lead you to the exact same simplified formula $\nu_{\text{beat}} = \frac{2 \nu_0 v_s}{v}$.

Question

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears $2	ext{ beats/s}$. The oscillation frequency of each tuning fork is $
u_0 = 1400	ext{ Hz}$ and the velocity of sound in air is $350	ext{ m/s}$. The speed of each tuning fork is close to

Options:

(A) $\frac{1}{8}	ext{ m/s}$

(B) $\frac{1}{2}	ext{ m/s}$

(C) $1	ext{ m/s}$

(D) $\frac{1}{4}	ext{ m/s}$

Sigma Percentile · Accepted Answer

### The Symphony of Motion

Imagine standing perfectly still in the center of a room. Suddenly, two identical tuning forks are set into motion. One glides smoothly towards you, while the other retreats at the exact same speed.

Even though both forks are identical and vibrate at the same natural frequency, the sound that reaches your ears is not a single, pure tone. Instead, you hear a rhythmic pulsing—a phenomenon known as **beats**.

This beautiful auditory illusion is a direct consequence of the **Doppler Effect**. Let's break down exactly why this happens and how we can use it to find the speed of the tuning forks.

### The Approaching Source

First, let's focus on the tuning fork moving towards you. As it travels, it "chases" its own sound waves. This compresses the waves in front of it, effectively shortening the wavelength.

Because the speed of sound in air remains constant, a shorter wavelength means a higher frequency reaches your ears. The formula for the apparent frequency $
u_1$ of an approaching source is:

$$ 
u_1 = \left(\frac{v}{v - v_s}ight)
u_0 $$

Here, $v$ is the speed of sound, $v_s$ is the speed of the source, and $
u_0$ is the original frequency. Notice that the denominator is smaller than $v$, which mathematically confirms that $
u_1 > 
u_0$.

### The Receding Source

Now, consider the second tuning fork moving away from you. The opposite effect occurs. The sound waves are stretched out behind the moving fork, leading to a longer wavelength and a lower frequency.

The formula for the apparent frequency $
u_2$ of a receding source is:

$$ 
u_2 = \left(\frac{v}{v + v_s}ight)
u_0 $$

In this case, the denominator is larger than $v$, confirming that $
u_2 < 
u_0$.

### The Birth of Beats

Because you are hearing both $
u_1$ and $
u_2$ simultaneously, the two sound waves interfere with each other. Since their frequencies are slightly different, they drift in and out of phase, creating a pulsating sound.

The frequency of this pulsation is called the **beat frequency** ($
u_{	ext{beat}}$), and it is simply the difference between the two apparent frequencies:

$$ 
u_{	ext{beat}} = 
u_1 - 
u_2 $$

Let's substitute our Doppler equations into this beat frequency formula:

$$ 
u_{	ext{beat}} = \left(\frac{v}{v - v_s}ight)
u_0 - \left(\frac{v}{v + v_s}ight)
u_0 $$

### Algebraic Elegance

To simplify this, we can factor out the common terms $v$ and $
u_0$:

$$ 
u_{	ext{beat}} = v 
u_0 \left[ \frac{1}{v - v_s} - \frac{1}{v + v_s} ight] $$

Taking a common denominator, we get:

$$ 
u_{	ext{beat}} = v 
u_0 \left[ \frac{(v + v_s) - (v - v_s)}{v^2 - v_s^2} ight] $$

The $v$ terms in the numerator cancel out, leaving $2v_s$:

$$ 
u_{	ext{beat}} = \frac{2 v 
u_0 v_s}{v^2 - v_s^2} $$

### The Crucial Approximation

Here is where we must pay close attention to the problem statement. We are told that the speed of the tuning forks is **much less than the speed of sound** ($v_s \ll v$).

Because $v_s$ is so small compared to $v$, squaring it makes it even more insignificant. Therefore, we can safely approximate the denominator:

$$ v^2 - v_s^2 \approx v^2 $$

Substituting this back into our equation yields a beautifully simple formula:

$$ 
u_{	ext{beat}} \approx \frac{2 v 
u_0 v_s}{v^2} = \frac{2 
u_0 v_s}{v} $$

### The Final Calculation

We are now ready to plug in the given values. We know the beat frequency $
u_{	ext{beat}} = 2 	ext{ Hz}$, the original frequency $
u_0 = 1400 	ext{ Hz}$, and the speed of sound $v = 350 	ext{ m/s}$.

$$ 2 = \frac{2 	imes 1400 	imes v_s}{350} $$

We can cancel the $2$ on both sides:

$$ 1 = \frac{1400}{350} v_s $$

Since $1400$ divided by $350$ is exactly $4$, we have:

$$ 1 = 4 v_s $$

Solving for $v_s$, we find:

$$ v_s = \frac{1}{4} 	ext{ m/s} $$

The tuning forks are moving at a gentle speed of $0.25 	ext{ m/s}$. The physics of the Doppler effect, combined with a smart mathematical approximation, leads us flawlessly to the correct answer!

Select Answer:

Visualized Solution

Visualizing the Setup $v_{s}, v, ν_{0}$

Approaching Frequency $ν_{1}$

Receding Frequency $ν_{2}$

Beat Frequency $ν_{beat}$

Substituting Frequencies

Algebraic Simplification

Applying Approximation $v_{s} ≪ v$

Substituting Values

Final Answer $v_{s}$

The Way Forward

The Sigma Insight: Doppler Effect

The Symphony of Motion

The Approaching Source

The Receding Source

The Birth of Beats

Algebraic Elegance

The Crucial Approximation

The Final Calculation

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