Animated Solution for Physics - Waves: A whistle producing sound waves of frequencies $9500\text{ Hz}$ and above is approaching a stationary person with speed $v{\text{ ms}}^{-1}$. The velocity of sound in air is $300{\text{ ms}}^{-1}$. If the person can hear frequencies upto a maximum of $10000\text{ Hz}$, the maximum value of $v$ upto which he can hear the whistle is

Understanding the Doppler Effect

Imagine standing on a quiet train platform. In the distance, you hear the whistle of an approaching train. As it gets closer, the pitch of the whistle seems to rise, becoming sharper and more urgent. Then, the moment it passes you, the pitch suddenly drops, transforming into a lower, more relaxed tone.

This fascinating phenomenon, where the perceived frequency of a wave changes due to the relative motion between the source and the observer, is known as the Doppler Effect. It's not just limited to sound; it applies to all types of waves, including light! In fact, astronomers use the Doppler effect of light to determine whether galaxies are moving towards or away from us.

In our problem, we are dealing with a classic Doppler effect scenario involving sound waves. Let's break down the physical setup and see how the math perfectly models reality.

We have a stationary person acting as our observer. This means the velocity of the observer, $v_o$, is exactly zero.

Approaching this person is a whistle—our sound source. The whistle is emitting sound waves at a true frequency of $f=9500\text{ Hz}$. It's moving towards the observer with an unknown speed, which we'll call $v_s=v$.

The problem states that the person can hear frequencies up to a maximum of $10000\text{ Hz}$. This is our apparent frequency, $f_{app}$. The speed of sound in the air is given as $v_{sound}=300{\text{ ms}}^{-1}$.

Our goal is to find the maximum speed $v$ at which the whistle can approach so that the person can still just barely hear it at $10000\text{ Hz}$.

To connect all these variables, we rely on the general formula for the Doppler effect:

This equation might look a bit intimidating with its plus-minus signs, but it's actually very logical. Let's tailor it to our specific situation.

First, since our observer is stationary, $v_o=0$. The numerator simply becomes $v_{sound}$.

Second, we need to decide on the sign in the denominator. The source is approaching the observer. When a source approaches, it "catches up" to the sound waves it just emitted, compressing the wavefronts together. This compression leads to a higher frequency reaching the observer.

For the apparent frequency $f_{app}$ to be greater than the true frequency $f$, the fraction must be greater than 1. This means the denominator must be smaller than the numerator. Therefore, we must use the negative sign in the denominator.

Our tailored equation becomes:

Now, let's plug in the values we know. We substitute $f_{app}=10000$, $f=9500$, $v_{sound}=300$, and $v_s=v$:

This is our raw mathematical setup. The physics part is done; now it's time for some algebra. Let's start by simplifying the equation. We can divide both sides by $9500$:

Canceling out the zeros, we get a much cleaner fraction on the left:

To get rid of the fractions, we perform cross-multiplication. We multiply the numerator of the left side by the denominator of the right side, and vice versa:

Now, let's isolate the term containing our unknown velocity, $v$. We can divide both sides of the equation by $100$:

The $300$ divided by $100$ simply becomes $3$. This leaves us with:

Calculating the product on the right side:

We are almost there! To find $v$, we just need to rearrange the terms. Let's move $v$ to the right side and $285$ to the left side:

Subtracting the numbers gives us our final answer:

And there we have it! The maximum speed at which the whistle can approach the person, while still allowing them to hear the sound at $10000\text{ Hz}$, is $15{\text{ ms}}^{-1}$.

This problem beautifully illustrates how the Doppler effect formula can be used to determine the speed of a moving object just by analyzing the shift in the frequency of the sound it emits. It's a powerful tool that bridges the gap between abstract wave mechanics and real-world observations.