Solving Frequency Problems: Motorists & Guitar Strings

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madeinmgs
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Alright so I've been studying up on frequency. Things like sound waves and a small area of the Doppler Effect. The problems envolve finding the frequency.

1.) Two motorists are traveling at 90.0 km/h in the same direction on a freeway, and each is equipped with a horn, that has frequency of 350 Hz.

(a) If the from motorist sounds her horn, what frequency does the near motorist hear?

(b) If the rear motorist sounds his horn, what frequency does the front motorist hear?

I understand that to find frequency, you take the speed of sound, depending on whether they are in front or behind each other, subtract it from the velocity of the object, and then multiply it by the frequency of the object making the sound. f' = (v - vo) / (v - vs) * fo

But the problem doesn't tell me that one is in front of the other, do I assume they are driving side by side, and if so, does the velocity subtract or add?


A 65.0 cm guitar string has a mass of 2.60 g and is plucked so as to produce its fundamental frequency. The sound wave emitted has a wavelength of the 1.17 m. What is the tension in the string?

This problem doesn't deal with harmonics ie.(open pipe, or closed pipe) I'm really stumped on this one. I can see trying to tackle this problem from a v = sqrt(F / u ) point. Which I think turns into a T(tension) = uV^2 But any push in the right direction would be helpful.
 
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in problem 1, one car is ahead of the other. they are going the same speed, so the distance between them is not changing, and that distance does not matter for the problem.

in problem 2, you have the right idea, but there are a couple more steps before the solution can be worked out. one thing to note is that you are given information about two different waves, one in the string and one in the air. the only thing they must have in common is their frequency. that frequency can be worked out from the sound wave. it can then be used to find the speed of the wave in the string. then, well, you have it worked out from there.

cheers
 
Well with the velocity of both vehicles being the same, what would be the point of dividing them? The speed of sound is 350m/s, you subtract that by the speed of the cars. v-vo / v-vs but both vo and vs are the same.
 
the problem is different depending on which car you are talking about. from the front cars perspective, it is the observer, and it is moving away from the source (the other car). but the source is moving toward the observer. so the ratio should be v-vo / v+vs .