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Standing wave (continue)

Hi again :biggrin:
The first part of my question was stated under "Simple math problem". However, the rest subproblems are no more simple math ones only :rolleyes:
I'm given the wave function [tex]y(x,t)=4.2\cos(0.2x)\sin(300t)[/tex].
(a) What are the speed and amplitude of the two travelling waves, that result in this standing wave?
(b) What is the distance between successive nodes on the string?
(c)If the string is vibrating in its fourth harmonic, how long is it?
About (a): I think that the two speeds are equal in magnitude and opposite in direction: [tex]v=f\lambda=\pm1500[/tex], and the amplitudes are half of that of the standing wave -> [tex]2.1[/tex]. :cool: Up to here I hope everything is correct...
About (b) I'm a little bit confused... What I think is that I should take the first derivative of the y(x,t) w.r.t. time and find its zero points (because the "particles" in the node do not move). Finding the difference between two successive "zeros" will give me the result. Is this correct?
About (c): honestly I don't have the slightest idea :uhh: .We didn't even study this in class?!?! why would the professor give us such a problem? :frown: ... But what ever... please, guys, check if the above ideas are correct and tell me the result.
Thank you in advance! :approve:
 
I just found out some clues and answers:
For (c): if it is vibrating in its fourth harmonic, the length is [tex]L=2\lambda=62,8[/tex].
For (b): The dist. between successive nodes is just half of the wavelength -> 15,7.
Does this mean that I answered myself? :biggrin: Where are you, guys?!?! Anybody home? :eek:
 

OlderDan

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Ramses The Pharaoh said:
I just found out some clues and answers:
For (c): if it is vibrating in its fourth harmonic, the length is [tex]L=2\lambda=62,8[/tex].
For (b): The dist. between successive nodes is just half of the wavelength -> 15,7.
Does this mean that I answered myself? :biggrin: Where are you, guys?!?! Anybody home? :eek:
Looks good to me.
 
Last edited:

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