# Speed of a wave on a string and Frequency of 3rd Harmonic

• ddsMom
In summary, the string has a speed of 310m/s when tension is 400 N and when frequency is 3rd harmonic.
ddsMom

## Homework Statement

Tension = 400 N
Mass = 4g
Length = .96m
What is the speed of the wave on a string?

What is the frequency of the 3rd harmonic?

v=√T/(m/L)

v=fλ

## The Attempt at a Solution

v=√400N/(.004kg/.96m) = 310m/s...am I correct?

f=v/λ
2nd harmonic f=v/(λ/2) = 310/(.96/2) = 646Hz

3rd harmonic f=v/(λ*2/3) = 646/.64 = 1009Hz

Are my answers correct?

Thank you so much for your input. Really trying to learn this!

ddsMom said:

## Homework Statement

Tension = 400 N
Mass = 4g
Length = .96m
What is the speed of the wave on a string?

What is the frequency of the 3rd harmonic?

v=√T/(m/L)

v=fλ

## The Attempt at a Solution

v=√400N/(.004kg/.96m) = 310m/s...am I correct?
Yes.

f=v/λ
2nd harmonic f=v/(λ/2) = 310/(.96/2) = 646Hz

3rd harmonic f=v/(λ*2/3) = 646/.64 = 1009Hz
What's ##\lambda##? I think you meant L. I got a different answer than you did for the frequency of the third harmonic.

1 person
I think I see my mistake. It should read...

f=v/λ
2nd harmonic f=v/(λ/2) = 310/(.96/2) = 646Hz
3rd harmonic f=v/(λ*2/3) = 310/.64 = 484Hz

Is that correct?

No. The frequency should increase with higher harmonics.

1 person
That's what I thought. What am I doing wrong?

3rd harmonic f=v/(λ*2/3) = 310/(.96/(2/3)) = 484Hz is my initial formula

Actually, your frequency for the third harmonic is right. The wavelength and therefore the frequency you have for the second harmonic is wrong.

Every time you go up one harmonic, you fit in another half wavelength on the length of the string. So for the first harmonic, you have ##1\times\lambda/2 = L##; for the second, you get ##2\times\lambda/2 = L##; and so on.

I got it. The frequency of the second harmonic is two times the frequency of the first harmonic. The frequency of the third harmonic is three times the frequency of the first harmonic.

so the answer is 969...I was going about it the hard way.

Thank you for your time!

Nooooo!

Really?

Yes, really. The frequencies aren't simply integer multiples of the fundamental frequency. You were on the right track earlier, but you were making some mistakes.

That would be 310/(3*λ/2) = 310/1.44 = 215.

That doesn't seem correct?

It's been over 20 years since I've had Physics so I definitely appreciate any help.

## 1. What is the relationship between the speed of a wave on a string and the frequency of the 3rd harmonic?

The speed of a wave on a string is directly proportional to the frequency of the 3rd harmonic. This means that as the frequency of the 3rd harmonic increases, the speed of the wave on the string also increases.

## 2. How does the length of the string affect the speed of a wave?

The length of the string has a direct impact on the speed of a wave. As the length of the string increases, the speed of the wave decreases. Similarly, as the length of the string decreases, the speed of the wave increases.

## 3. Can the speed of a wave on a string be changed?

Yes, the speed of a wave on a string can be changed by altering the tension of the string or by changing the properties of the medium through which the wave is traveling.

## 4. What is the formula for calculating the speed of a wave on a string?

The formula for calculating the speed of a wave on a string is v = √(T/μ), where v is the speed of the wave, T is the tension of the string, and μ is the mass per unit length of the string.

## 5. How does the frequency of the 3rd harmonic compare to the fundamental frequency?

The frequency of the 3rd harmonic is three times the frequency of the fundamental frequency. In other words, if the fundamental frequency is f, the frequency of the 3rd harmonic would be 3f.

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