An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of

Q: Why do we use a plus sign in the numerator for the observer's velocity?

When you move towards a sound source, you are essentially 'rushing into' the sound waves. This means you cross more wave crests per second than you would if you were standing still. A higher rate of crossing crests translates to a higher perceived frequency, which is mathematically represented by adding your velocity to the wave's velocity in the numerator.

Q: Does the actual speed of sound ($v$) affect the final percentage?

No, it doesn't! As you can see in the calculation, the speed of sound $v$ appears in both the numerator and denominator and perfectly cancels out. The percentage change depends entirely on the *ratio* of the observer's speed to the speed of sound, not their absolute values.

Q: What if the source was moving towards the observer instead?

If the source were moving towards a stationary observer, the waves would get 'bunched up' in front of the source, effectively decreasing the wavelength. In the Doppler formula, this is represented by subtracting the source's velocity in the denominator: $f' = f \left( \frac{v}{v - v_s} \right)$.

Q: Why is the percentage increase calculated as $\frac{f' - f}{f}$ and not just $f'$?

The apparent frequency $f'$ tells you the *new* frequency you hear. But the question asks for the *increase*. The increase is the difference between the new and old frequency ($f' - f$). To find the *percentage* increase, we divide this difference by the original frequency $f$ and multiply by 100.

Q: Can the apparent frequency ever be lower than the original frequency?

Absolutely! If the observer were moving *away* from the source, they would be 'running away' from the wave crests, crossing them less frequently. In that case, we would use a minus sign in the numerator ($v - v_o$), resulting in an apparent frequency lower than the original.

Question

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?

Options:

(A) Zero

(B) 0.5%

(C) 5%

(D) 20%

Sigma Percentile · Accepted Answer

### The Symphony of Sound and Motion

Imagine you are standing on a quiet street when a police car with its siren blaring speeds towards you. You have probably noticed that the pitch of the siren sounds higher as it approaches and lower as it speeds away. This everyday phenomenon is the **Doppler Effect**.

In this problem, we are flipping the script. The source of the sound is perfectly stationary, and *you* (the observer) are the one moving towards it. Because you are rushing headlong into the oncoming sound waves, you are crossing the wave crests more frequently than if you were just standing still. This increased rate of interception translates directly into a higher perceived frequency.

### Setting Up the Stage

Let's define our physical parameters clearly to avoid any confusion. We are given that the source is stationary, which means its velocity is zero ($v_s = 0$).

The observer is moving towards the source with a velocity that is one-fifth the speed of sound. If we let $v$ represent the speed of sound in the medium, then the observer's velocity is:

$$v_o = \frac{v}{5}$$

### The Master Equation

To find out exactly what frequency the observer hears, we rely on the general Doppler effect formula. The formula relates the apparent frequency ($f'$) to the original frequency ($f$) based on the relative velocities of the source and observer:

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

Since our source is stationary, the denominator simply becomes $v$. Because the observer is moving *towards* the source (increasing the frequency), we must use the **plus sign** in the numerator. Our master equation simplifies beautifully to:

$$f' = f \left( \frac{v + v_o}{v} ight)$$

### Executing the Math

Now, let's substitute the given value of $v_o$ into our simplified equation. Notice how keeping everything in terms of $v$ makes the algebra incredibly clean:

$$f' = f \left( \frac{v + \frac{v}{5}}{v} ight)$$

To simplify the numerator, we find a common denominator:

$$v + \frac{v}{5} = \frac{5v}{5} + \frac{v}{5} = \frac{6v}{5}$$

Plugging this back into our frequency equation, the $v$ in the numerator and denominator cancel out perfectly:

$$f' = f \left( \frac{\frac{6v}{5}}{v} ight) = \frac{6}{5}f$$

This tells us that the apparent frequency is $\frac{6}{5}$ times the original frequency.

### The Final Crescendo

The question doesn't just ask for the new frequency; it asks for the **percentage increase**. The formula for percentage increase is the change in value divided by the original value, multiplied by 100:

$$\% 	ext{ increase} = \left( \frac{f' - f}{f} ight) 	imes 100$$

Let's substitute our result for $f'$:

$$\% 	ext{ increase} = \left( \frac{\frac{6}{5}f - f}{f} ight) 	imes 100$$

Factoring out $f$ in the numerator:

$$\% 	ext{ increase} = \left( \frac{f(\frac{6}{5} - 1)}{f} ight) 	imes 100$$

The $f$ cancels out, leaving us with:

$$\% 	ext{ increase} = \left( \frac{1}{5} ight) 	imes 100 = 20\%$$

And there we have it! By moving towards the source at one-fifth the speed of sound, the observer experiences a crisp, exact **20% increase** in the apparent frequency.

### Beyond the Problem

What if the observer had been running *away* from the source at the same speed? The physics flips! You would be running away from the wave crests, crossing them less frequently. Mathematically, the sign in the numerator of our Doppler formula would change to a minus ($v - v_o$), resulting in an apparent frequency of $\frac{4}{5}f$, which is a 20% *decrease*. Always let the physical reality guide your mathematical signs!

Animated Solution for Physics - Waves: 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?

Select Answer:

Visualized Solution

\text{Visualizing the Scenario}

\text{Doppler Effect Formula}

\text{Substituting the Values}

\text{Calculating Apparent Frequency}

\text{Percentage Increase}

\text{The Way Forward}

The Sigma Insight: Doppler Effect

The Symphony of Sound and Motion

Setting Up the Stage

The Master Equation

Executing the Math

The Final Crescendo

Beyond the Problem

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