Physics Periodic Motion and Waves Questions (NOVA)

[osraimingover40]

Hi guys,
I don't quite understand this problem. " The wave velocity is given by v= sqrt (T/ u) .... For the following questions, consider an E string (frequency 660 Hz) which is made of steel. It has a mass of 0.66 grams for each meter of wire and has a circular cross section of diameter 0.33 mm. The string length when strung on a guitar is 0.65 m. Also note that the D string has a wave velocity of 382 m/s."

If the guitarist places her left finger lightly on the D string one fourth way from the neck end to the base, what is the lowest frequency that will be heard?

A) 588 Hz
B) 882 Hz
C) 1175 Hz
D) 2350 Hz


I don't quite understand why the answer is C. If its placed 1/4 through wouldn't it mean the L is only 3/4, not that the length if 1/4 of the original? (Can someone please explain this whole node thing to me?)


And
#6) If we want to increase the frequency of the fundamental of a string by 3%, by how much do we want to change the tension in the string?

So since v=f*wavelength, and v= sqrt (T/u) .... we can set f=sqrt(T/u) * wavelength ....so T would actually need to increase by f squared, which means roughy 6% correct?



Thank you.

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[Yorick]

Solution:

The wavelength of the string is given by 2L=2(.65)= 1.3m. f=v/(lambda) = 383/1.3= 293.8 Hz. However, the guitarist strung one fourth the way up the neck so the lowest heard frequency is the fourth harmonic and f_n=nf_1 = 4(293.8)= 1175Hz ->C Hope this helps.

Solution:

v=f(lambda)= sqrt(T/u) so f=sqrt (T/u) / (lambda). Increasing the frequency by 3% means we multiply f by 1.03 so assuming u and lambda are unchanged and letting t be the new tension, 1.03f = sqrt(t/u) / lambda substituting from earlier, 1.03 (sqrt (T/u) / (lambda)) = sqrt(t/u) / lambda , 1.0609T=t so we increase the tension by 6.09 percent, so yes, approximately 6%.