A boat is travelling in a river with a speed 10 m/s along the stream flowing with a speed 2 m/s. Fro

Q: Why does the river flow velocity add to the speed of sound in part (a)?

Because the sound waves are propagating downstream, in the same direction as the river flow. The moving water acts as a moving medium, carrying the sound waves along with it and increasing their effective speed relative to the ground.

Q: Why does the wind velocity subtract from the speed of sound in part (b)?

The wind is blowing in the direction opposite to the river stream, which means it is blowing against the direction of sound propagation (from the boat to the downstream receiver). This opposes the wave motion, reducing the effective speed of sound.

Q: Does the natural frequency of the transmitter change when it is moved from water to air?

No, the natural frequency of the transmitter is a characteristic of the source itself and remains constant at $10^5\text{ Hz}$ regardless of the medium.

Q: Why is the frequency shift much larger in air than in water?

The Doppler shift depends on the ratio of the source speed to the speed of sound ($v_s / v$). Since the speed of sound in air ($344\text{ m/s}$) is much smaller than in water ($1445\text{ m/s}$), the same source speed of $10\text{ m/s}$ produces a much larger relative shift in air.

Q: What temperature scale should be used in the speed of sound formula for air?

Always use the absolute temperature scale in Kelvin ($K$). Convert Celsius to Kelvin by adding $273.15$ (or $273$ for approximation).

Question

A boat is travelling in a river with a speed $10	ext{ m/s}$ along the stream flowing with a speed $2	ext{ m/s}$. From this boat a sound transmitter is lowered into the river through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is $14.45	ext{ mm}$. Assume that attenuation of sound in water and air is negligible.

(a) What will be the frequency detected by a receiver kept inside the river downstream?

(b) The transmitter and the receiver are now pulled up into air. The air is blowing with a speed $5	ext{ m/s}$ in the direction opposite to the river stream. Determine the frequency of the sound detected by the receiver.

(Temperature of the air and water $= 20^\circ	ext{ C}$ ; Density of river water $= 10^3	ext{ kg/m}^3$ ; Bulk modulus of the water $= 2.088 	imes 10^9	ext{ Pa}$ ; Gas constant $R = 8.31	ext{ J/mol-K}$ ; Mean molecular mass of air $= 28.8 	imes 10^{-3}	ext{ kg/mol}$ ; $\frac{C_p}{C_v}$ for air $= 1.4$)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on the banks of a flowing river.

A boat glides downstream, carrying a sound transmitter that hums at a constant frequency.

Downstream, a stationary receiver waits to capture these acoustic waves.

This classic problem from **JEE Advanced 2001** beautifully combines fluid dynamics, thermodynamics, and wave mechanics through the lens of the **Doppler Effect**.

Let's break down the physics step-by-step.

---

### Speed of Sound in Water

Before we can analyze the frequency shift, we must determine how fast sound travels through the river water.

Unlike air, water is highly incompressible, which means its bulk modulus $B$ is extremely high.

The speed of sound in a liquid medium is given by the formula:

$$v_w = \sqrt{\frac{B}{ho}}$$

Substituting the given bulk modulus $B = 2.088 	imes 10^9	ext{ Pa}$ and the density of water $ho = 10^3	ext{ kg/m}^3$:

$$v_w = \sqrt{\frac{2.088 	imes 10^9}{10^3}} = \sqrt{2.088 	imes 10^6} = 1445	ext{ m/s}$$

This speed is nearly four times faster than the speed of sound in air, showcasing how efficiently tightly packed water molecules transmit mechanical vibrations.

---

### Finding the Natural Frequency

We are given that the wavelength of the sound emitted inside the water is $\lambda_w = 14.45	ext{ mm}$.

Using the wave equation, we can find the natural frequency $f_0$ of the transmitter:

$$f_0 = \frac{v_w}{\lambda_w}$$

Substituting our calculated speed of sound and converting the wavelength to meters ($14.45 	imes 10^{-3}	ext{ m}$):

$$f_0 = \frac{1445}{14.45 	imes 10^{-3}} = 10^5	ext{ Hz}$$

This is a high-frequency ultrasonic wave, well above the human hearing range.

---

### Part (a): Doppler Effect in a Moving Medium

Now, let's address the motion of the medium.

The river is flowing downstream at $v_m = 2	ext{ m/s}$.

Because the medium itself is moving in the direction of sound propagation (towards the receiver), it carries the sound waves along with it.

Thus, the effective speed of sound relative to the ground is:

$$v_{	ext{eff}} = v_w + v_m = 1445 + 2 = 1447	ext{ m/s}$$

The boat (source) is moving downstream at $v_s = 10	ext{ m/s}$ relative to the ground, chasing its own sound waves.

The receiver is stationary ($v_o = 0$).

Applying the general Doppler formula:

$$f_1 = f_0 \left( \frac{v_{	ext{eff}} - v_o}{v_{	ext{eff}} - v_s} ight)$$

Substituting our values:

$$f_1 = 10^5 \left( \frac{1447 - 0}{1447 - 10} ight) = 10^5 \left( \frac{1447}{1437} ight) \approx 1.0069 	imes 10^5	ext{ Hz}$$

Because the source is moving towards the receiver, the detected frequency is slightly higher than the natural frequency.

---

### Part (b): Transitioning to Air with Wind

In the second part of the problem, the entire apparatus is pulled up into the air.

First, we must calculate the speed of sound in air $v_a$ at $20^\circ	ext{ C}$ ($293	ext{ K}$):

$$v_a = \sqrt{\frac{\gamma RT}{M}}$$

Substituting the given values:

$$v_a = \sqrt{\frac{1.4 	imes 8.31 	imes 293}{28.8 	imes 10^{-3}}} \approx 344	ext{ m/s}$$

Now, we are told that the wind is blowing at $w = 5	ext{ m/s}$ in the direction opposite to the river stream.

Since the sound is propagating downstream (towards the receiver), the wind is blowing directly against the sound waves.

This reduces the effective speed of sound in air to:

$$v_{	ext{eff}} = v_a - w = 344 - 5 = 339	ext{ m/s}$$

Applying the Doppler formula again with this new effective speed:

$$f_2 = f_0 \left( \frac{v_{	ext{eff}} - v_o}{v_{	ext{eff}} - v_s} ight)$$

$$f_2 = 10^5 \left( \frac{339 - 0}{339 - 10} ight) = 10^5 \left( \frac{339}{329} ight) \approx 1.0304 	imes 10^5	ext{ Hz}$$

Notice how the frequency shift is more pronounced in air than in water.

This is because the source speed ($10	ext{ m/s}$) is a much larger fraction of the speed of sound in air ($344	ext{ m/s}$) than in water ($1445	ext{ m/s}$).

This elegant result highlights the deep connection between the medium's properties and wave propagation dynamics.

Visualized Solution

Visualizing the Setup

Calculating the Speed of Sound in Water

Substituting Values for $v_{w}$

Finding the Natural Frequency of the Transmitter

Substituting Values for $f_{0}$

Doppler Effect with Moving Medium (Water)

Calculating $f_{1}$ in Water

Speed of Sound in Air

Substituting Values for $v_{a}$

Doppler Effect with Wind in Air

Calculating $f_{2}$ in Air

The Sigma Insight: Doppler Effect

Analyzing the Setup

Speed of Sound in Water

Finding the Natural Frequency

Part (a)

Part (b)

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