Solving Frequency Problems: Motorists & Guitar Strings

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In summary: in problem 1, the front car's sound wave has a shorter wavelength, so the tension in the string is lower.in problem 2, the string has the same frequency as the sound waves from the air and the rear car, so the tension is the same.
  • #1
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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  • #2
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
 
  • #3
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.
 
  • #4
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 .
 
  • #5


I would like to provide some guidance and clarification on your understanding of frequency problems. Firstly, frequency is a measure of how many oscillations or vibrations occur in a given time period. In the case of sound waves, it is measured in hertz (Hz), which represents the number of oscillations per second.

In the first problem, the frequency of the horn is given as 350 Hz, which means that the horn produces 350 vibrations per second. When one motorist sounds the horn, the sound waves travel towards the other motorist. If the two motorists are driving side by side, the velocity of the sound wave relative to the other motorist is zero. In this case, the frequency heard by the other motorist would simply be 350 Hz, as there is no change in the relative velocity between the sound wave and the other motorist.

If the rear motorist sounds the horn, the sound waves would travel towards the front motorist with a velocity of 90.0 km/h. In this case, the frequency heard by the front motorist would be slightly higher than 350 Hz, as the sound waves are approaching the front motorist. This phenomenon is known as the Doppler effect, where the frequency of a sound wave changes based on the relative motion between the source and the observer.

In the second problem, the fundamental frequency of a guitar string is the lowest frequency at which the string can vibrate. It is given by the equation f = 1/2L * sqrt(T/μ), where L is the length of the string, T is the tension in the string, and μ is the linear density of the string. In this case, the linear density is given by μ = m/L, where m is the mass of the string and L is its length. By substituting the given values and solving for T, you can find the tension in the string.

It is important to note that the fundamental frequency of a string is not affected by whether it is an open or closed pipe, as this concept only applies to the harmonics of a string. I hope this helps to clarify your understanding of frequency problems. Keep exploring and learning about this fascinating subject!
 

Related to Solving Frequency Problems: Motorists & Guitar Strings

1. How do you solve frequency problems involving motorists and guitar strings?

To solve frequency problems involving motorists and guitar strings, you need to first understand the relationship between frequency and speed. The frequency of a sound wave is directly proportional to the speed of the object producing the sound. This means that as the speed of an object, such as a car, increases, the frequency of the sound produced by its engine also increases. Similarly, the frequency of a guitar string is directly related to the speed at which it vibrates. By understanding this relationship, you can use the formula f=NV/L to solve frequency problems involving motorists and guitar strings, where N is the number of harmonics, V is the speed of the object or string, and L is the length of the string.

2. What is the significance of solving frequency problems involving motorists and guitar strings?

Solving frequency problems involving motorists and guitar strings can provide valuable information about the speed of an object or the properties of a guitar string. For example, by calculating the frequency of a car's engine sound, you can determine how fast the car is moving. Similarly, by solving for the frequency of a guitar string, you can determine its tension, length, or mass. This knowledge can be useful in various fields such as physics, engineering, and music.

3. Can the formula f=NV/L be used for any type of object or string?

Yes, the formula f=NV/L can be used for any type of object or string as long as the relationship between frequency and speed applies. This formula is based on the fundamental frequency of a vibrating string, which is the frequency at which the string vibrates when it is in its lowest energy state. This concept can be applied to any object or string that produces sound waves.

4. What are some real-life applications of solving frequency problems involving motorists and guitar strings?

Solving frequency problems involving motorists and guitar strings has many real-life applications. For instance, it can help in designing and tuning musical instruments, analyzing the motion and speed of moving objects, and understanding the properties of sound waves. It can also be used in fields such as acoustics, mechanics, and signal processing.

5. Are there any limitations to solving frequency problems involving motorists and guitar strings?

Like any mathematical formula, the formula f=NV/L has its limitations. It assumes that the object or string is vibrating at its fundamental frequency, which may not always be the case. In addition, it does not take into account external factors such as air resistance or friction, which can affect the speed of an object or the vibrations of a string. Therefore, it is important to consider these factors and make necessary adjustments when using this formula to solve frequency problems involving motorists and guitar strings.

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