How Fast is the Racecar Moving Based on the Doppler Effect?

[BlueSkyy]

Homework Statement

The pitch of the sound from a racecar engine drops the musical interval of a fourth when it passes the spectators. This means the frequency of the sound after passing is 0.75 times what it was before. How fast is the racecar moving?
1. 61 m/s
2. 49 m/s
3. 72 m/s
4. 68 m/s

Homework Equations

fo = fs / 1 - (vs/v)

The Attempt at a Solution

I tried using 75 = 100 / (1 - (vs/331) to solve for the speed of the source, but my answer was way off...there's got to be something I'm missing. am i correct in thinking i need to use v = 331 for the speed of sound? then what do i use for fo and fs?

 

[G01]

Your problem is that fs is not 100.

100 is the observed frequency for when the car is moving toward you.

Then 75, would be the observed frequency for when the car is moving away from you.

So you are going to have to set up two doppler shift equations, one for when the car is moving toward and the other for when the car is moving away from you.

So, after you have these two equations, how would you solve for v?

 

[ogb p]

I would approach this problem by first understanding the concept of the Doppler effect. The Doppler effect is the change in frequency of a wave (in this case, sound) due to the relative motion between the source of the wave and the observer. In this case, the source of the sound is the racecar and the observer is the spectator.

The formula provided (fo = fs / 1 - (vs/v)) is correct, but it is important to note that the speed of sound (v) is not a constant value. It depends on various factors such as temperature, humidity, and altitude. In this case, we can assume that the speed of sound is approximately 331 m/s, as this is the speed of sound at sea level and room temperature.

To solve for the speed of the racecar (vs), we need to know the original frequency of the sound (fo) and the frequency of the sound after passing the spectator (fs). The musical interval of a fourth means that the frequency drops by a factor of 0.75, so we can set up the equation as:

0.75 = fo / fs

We can rearrange this to solve for fs:

fs = fo / 0.75

Substituting this into the original equation, we get:

fo = (fo / 0.75) / (1 - (vs/331))

Simplifying, we get:

0.75 = 1 - (vs/331)

Rearranging and solving for vs, we get:

vs = (1 - 0.75) * 331

vs = 0.25 * 331

vs = 82.75 m/s

Therefore, the speed of the racecar is approximately 82.75 m/s. This is option 4, 68 m/s, rounded to the nearest whole number.

In conclusion, to solve this problem, we needed to understand the concept of the Doppler effect and use the provided formula, making sure to account for the speed of sound not being a constant value.