A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its fr

Q: Why is there a minus sign in the denominator of the Doppler formula?

When the source moves towards the observer, it 'catches up' to the sound waves it just emitted. This compresses the waves, decreasing the wavelength and increasing the frequency. The minus sign in the denominator makes the overall fraction greater than 1, reflecting this increase.

Q: What if the train was moving away from the observer?

If the train were moving away, the waves would be stretched out, leading to a lower frequency. In that case, the formula would have a plus sign in the denominator: $f_{obs} = f \left( \frac{v}{v + v_s} \right)$.

Q: Do we need to know the original frequency of the whistle?

No, we don't! Because the problem asks for the *ratio* of the two observed frequencies, the original frequency $f$ appears in both the numerator and the denominator and cancels out completely.

Q: Why shouldn't I convert the fractions to decimals right away?

Converting to decimals early on can lead to messy calculations and rounding errors. By keeping the expressions as fractions, you allow common terms (like 340 and $f$) to cancel out, making the final arithmetic much simpler and exact.

Q: Is the speed of sound always $340 \text{ m/s}$?

The speed of sound depends on the medium and its temperature. In air at room temperature, it's approximately $343 \text{ m/s}$. However, in physics problems, $340 \text{ m/s}$ or $330 \text{ m/s}$ is often given to make calculations easier. Always use the value provided in the question!

Question

A train moves towards a stationary observer with speed $34	ext{ m/s}$. The train sounds a whistle and its frequency registered by the observer is $f_1$. If the speed of the train is reduced to $17	ext{ m/s}$, the frequency registered is $f_2$. If speed of sound is $340	ext{ m/s}$, then the ratio $f_1 / f_2$ is

Options:

(A) $\frac{19}{18}$

(B) $\frac{21}{20}$

(C) $\frac{20}{19}$

(D) $\frac{18}{17}$

Sigma Percentile · Accepted Answer

Imagine you are standing near a railway track. A train is approaching you, blowing its whistle. Because the train is moving towards you, the sound waves get compressed, and you hear a higher pitch. This is the classic **Doppler effect**.

### The Master Equation

To find the exact frequency you hear, we use the Doppler formula for a moving source and a stationary observer. The apparent frequency is given by:

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

Here, $f$ is the original frequency, $v$ is the speed of sound, and $v_s$ is the speed of the source. Notice the minus sign in the denominator—it makes the fraction greater than one, which mathematically explains why the frequency increases as the train approaches.

### Analyzing the Two Cases

Let's look at the first scenario. The train is moving at $34	ext{ m/s}$. We substitute this into our formula, along with the speed of sound, which is $340	ext{ m/s}$. We will call this observed frequency $f_1$.

$$f_1 = f \left( \frac{340}{340 - 34} ight)$$

Subtracting $34$ from $340$ gives us $306$. So, $f_1$ is $f$ times $340$ over $306$. Let's keep it as a fraction for now; it will make our calculations easier later.

$$f_1 = f \left( \frac{340}{306} ight)$$

Now, the train slows down to $17	ext{ m/s}$. We set up our equation again for the new observed frequency, $f_2$. We plug in $17$ for the source speed.

$$f_2 = f \left( \frac{340}{340 - 17} ight)$$

$340$ minus $17$ is $323$. So, $f_2$ becomes $f$ times $340$ over $323$.

$$f_2 = f \left( \frac{340}{323} ight)$$

### Final Calculation

The question asks for the ratio of $f_1$ to $f_2$. Let's divide our two expressions. Notice how the original frequency $f$ and the speed of sound, $340$, are in both the numerator and the denominator.

$$\frac{f_1}{f_2} = \frac{f \left( \frac{340}{306} ight)}{f \left( \frac{340}{323} ight)}$$

The $f$ and the $340$ cancel out beautifully. We are left with $323$ divided by $306$.

$$\frac{f_1}{f_2} = \frac{323}{306}$$

Both numbers are divisible by $17$. $323$ is $19 	imes 17$, and $306$ is $18 	imes 17$. Canceling the $17$s, we get our final ratio:

$$\frac{f_1}{f_2} = \frac{19}{18}$$

This perfectly matches option (a).

Select Answer:

Visualized Solution

Doppler Effect Setup

Doppler Formula

Case 1: $v_{s} = 34 m/s$

Simplifying $f_{1}$

Case 2: $v_{s} = 17 m/s$

Simplifying $f_{2}$

Ratio $\frac{f _{1}}{f _{2}}$

Final Calculation

The Sigma Insight: Doppler Effect

The Master Equation

Analyzing the Two Cases

Final Calculation

Similar Questions

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 = $320 ms^{- 1}$ )

An observer standing on a railway crossing receives frequency of $2.2 kHz$ and $1.8 kHz$ when the train approaches and recedes from the observer. Find the velocity of the train. (The speed of the sound in air is $300 m/s$ .)

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 $= 300 m/s$ )

A whistle giving out $450 Hz$ approaches a stationary observer at a speed of $33 m/s$ . The frequency heard by the observer (in $Hz$ ) is (Speed of sound $= 330 m/s$ )

Select Answer:

Visualized Solution

Doppler Effect Setup

Doppler Formula

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vs​=34 m/s

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f1​

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vs​=17 m/s

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f2​

Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f2​f1​​

Final Calculation

The Sigma Insight: Doppler Effect

The Master Equation

Analyzing the Two Cases

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A whistle giving out 450 Hz approaches a stationary observer at a speed of 33 m/s. The frequency heard by the observer (in Hz) is (Speed of sound =330 m/s)

Case 1: $v_{s} = 34 m/s$

Simplifying $f_{1}$

Case 2: $v_{s} = 17 m/s$

Simplifying $f_{2}$

Ratio $\frac{f _{1}}{f _{2}}$

A whistle giving out $450 Hz$ approaches a stationary observer at a speed of $33 m/s$ . The frequency heard by the observer (in $Hz$ ) is (Speed of sound $= 330 m/s$ )