# String vibration and frequency,amplitude,length

1. Sep 12, 2007

### Leopold Stotch

A string vibrates at its third-harmonic frequency. The amplitude at a point 30 cm from one end is half the maximum amplitude. How long is the string?

i figured since the amplitude at point 30 cm is half the max amplitude, the amplitude must be at 60 cm. The third harmonic frequency means that there are 4 nodes and 3 antinodes. i keep getting 1.2 for the length of one but since there are 3, i multiplied 1.2 x 3 and got 3.6m but i don't think that answer can be right. thanks for the help!

2. Sep 12, 2007

### learningphysics

careful... you don't know the amplitude... the 30cm is not an amplitude... it is a position along the string...

The first step is to find the wavelength of this vibration.

This is a standing wave... what is the equation for a standing wave?

3. Sep 12, 2007

### Leopold Stotch

from my notes i believe that the wavelength of the string is 2/3 the total length

4. Sep 12, 2007

### learningphysics

exactly... you'll use that to get the length of the string... but you first need to find the wavelength. once you have the wavelength, then you know the length of the string is 3/2 the wavelength...

5. Sep 12, 2007

### Leopold Stotch

is there any way to find the frequency(number value) just by knowing it's the third harmonic?

6. Sep 12, 2007

### learningphysics

We don't need the frequency here. The equation for a standing wave is:

$$y = Acos({\omega}t)sin(kx)$$

But we only need the amplitudes at the different points x... ie: the max value... that occurs when cos(omega*t) = 1...

So the equation for amplitudes is:

s = Asin(kx), where A is the maximum amplitude

find k, and using k find the wavelength at which the string is vibrating...

7. Sep 12, 2007

### Leopold Stotch

i'm still not sure where to find the values for A and x. i'm sorry for being such a pain

8. Sep 12, 2007

### learningphysics

no prob. you're not being a pain at all. The idea is to substitute s = (1/2)A and x = 30, into the equation:

s = Asin(kx)

and then solve for k. the A's will cancel.

Last edited: Sep 12, 2007
9. Sep 12, 2007

### Leopold Stotch

ok i get 1 as an answer for k....

(inverse sin(.5))/30 = 1

10. Sep 12, 2007

### learningphysics

cool. yup, that's right... k = 1 degree/cm. Can you get the wavelength from this?

11. Sep 13, 2007

### Leopold Stotch

not sure exactly what to do next...

12. Sep 13, 2007

### learningphysics

Hint, one wavelength occurs when the angle changes by 360 degrees... at x = 0 then angle is kx = 0. So you're looking for the x where kx = 360... that gives the wavelength...

Be careful in general though... although we're using degrees here, usually k is given in radians/unit length... it's all good as long as you know what's going on.

13. Sep 13, 2007

### Leopold Stotch

so that would mean 360 cm is 1 wavelength. If it is the third harmonic, then L=(3/2)(3.6m)=5.4m?

14. Sep 13, 2007

### learningphysics

yup... that's what I get. there's an assumption we made in the problem... that the 30cm was the first time we get half the amplitude...

ie we went from sin(30k) = 1/2 to 30k = 30 degrees... technically... 150 degrees also works... so does 210 degrees... (sin(210) is -1/2 but the amplitude is still 1/2 A)...

we assumed the 30cm was at 30degrees... hope it is the right assumption... the question doesn't make it clear.

15. Sep 13, 2007

### Leopold Stotch

hmm darn i was told that 5.4m was wrong.... oh well, i appreciate the help, i'm going with the answers of either 2.25m or 3.15m, thank you very very much!

16. Sep 13, 2007

### learningphysics

oh, crap. sorry man! is it multiple choice or do you just enter the answer?

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