2 questions about waves (interference, resonance)

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
coconut62
Messages
161
Reaction score
1

Homework Statement



Please refer to the attached image.

For the first question,

The solution calculated the wavelength,λ by taking the whole length of the string (0.75m) as equal to 3λ/2.

What I don't understand is, since the string have to bend to produce a wave, how can we just take the displacement of the waves as the length of the string?


For question two,

What does it mean by "there is no resonant frequency between these two"? What significance does this sentence have?
 

Attachments

  • 20130519_124104.jpg
    20130519_124104.jpg
    43.2 KB · Views: 423
  • 20130519_131956.jpg
    20130519_131956.jpg
    39.9 KB · Views: 462
Physics news on Phys.org
you were given the length of the string. Each loop would correspond to1/2 lambda.

I think that no resonate freq means that 135 and 180 are not multiples of each other.
 
barryj said:
I think that no resonate freq means that 135 and 180 are not multiples of each other.
No, that's not what it means.
If the fundamental frequency is f, what are all the resonant frequencies? If there are no resonant frequencies between 135 and 180, what's their relationship in this list?
 
barryj said:
you were given the length of the string. Each loop would correspond to1/2 lambda.

Yes, but is the length of the curved part of the loop counted, or the "shortest distance from one end of the loop to another", which is a straight line?

haruspex said:
No, that's not what it means.
If the fundamental frequency is f, what are all the resonant frequencies?
If there are no resonant frequencies between 135 and 180, what's their relationship in this list?

First order, second order and so on?

Still don't understand :confused:
 
If you take a string, like a guitar string and pluck it in the middle of the string, the string will vibrate at the fundamental frequency. If you take the same guitar string and touch it in the middle with your finger and then pluck the string 1/4 from the end, it will vibrate at 2 times the fundamental frequency. Ifd you put your finger at 1/3 from the end and pluck it at 1/6 from the end, the string will vibrate at 3 times the fundamental frequency. This is a demonstration that the string can vibrate at integer multiples of the fundamental frrequency. Also notice that the difference between the harmonics of a string held at both ends is equal to the fundamental frequency. If the fundamental is say 100 Hz, then the harmonics will be 200 Hz, 300Hz, 400 Hz, etc. I hope this insite helps.