A fixed string in the third harmonic

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SUMMARY

The discussion focuses on calculating the amplitude of a vibrating string in its third harmonic, with a wave speed of 195 m/s and a frequency of 230 Hz. The amplitude at the antinode is given as 0.370 cm. Participants clarify the use of the harmonic formula and the relationship between wave speed, frequency, and wavelength, ultimately deriving the amplitude at a specific point on the string. The final calculations yield an amplitude of approximately 0.0036854 cm at a distance of 20 cm from the left end of the string.

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  • #31
Using:

v(x,t) = A_S_W \omega sin(\frac{\pi }{2}) cos(0)

gives me: 5.33

then

a(x,t) = A_S_W \omega ^2 sin(\frac{\pi }{2}) sin(\frac{\pi }{2})

gives me: 7700

Both of which are right. I disagree with mastering physics - if you diffentiate cos, it gives you minus sin:

a(x,t) = -A_S_W \omega ^2 sin(\frac{\pi }{2}) sin(\frac{\pi }{2}) giving -7700, which it said was wrong?

Many Thanks,

TFM
 
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  • #32
Indeed you do get minus sin, but -7700 would be the maximum deceleration, and it asks for the maximum acceleration. At least that's what I tell myself.

Thanks for your help with the period, I think I was having a mental bock.
 
  • #33
Have you got the right answer yet for the time, as I have left the thread marked unsolved until you have finished.:smile:

I think I was having a mental bock.

Don't worry, I think it happens to all of us from time to time!

TFM
 
  • #34
Yeah, I got in the end. Thanks for that :smile:.

By the way are you still working on the 'instantaneous power in a standing wave' problem? Sketching the graphs is quite tricky.
 

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