You are playing a note that has a fundamental frequency of 349.23 Hz on a guitar

Question

You are playing a note that has a fundamental frequency of 349.23 Hz on a guitar string of length 56.5 cm. At the same time, your friend plays a fundamental note on an open organ pipe, and 4.0 beats per seconds are heard. The mass per unit length of the string is 2.17 g/m. Assume the velocity of sound is 343 m/s. What are the possible frequencies of the open organ pipe? Enter the lower frequency first. When the guitar string is tightened, the beat frequency decreases. Find the original tension in the string. What is the length of the organ pipe?

Explanation / Answer

(a) f_beat=f2-f1

f2= frequency of open pipe,

f1= frequency of guitar,

f2 = f-beat+f1=4+349.23 =353.23 Hz.

(b)

frequency,

f1=v/2L=root(T/m/L) /2L

4f^2L^2=T*L/m

Tension in the string,

T=4*f^2*L*m =4*(349.23)^2*56.5*10^-2*2.17*10^-3

=598 N

(c)

For organ pipe, (L')

f2=v/2L'=root(T/m/L') /2L'

4f2^2L^2=T*L/m

L=(T/m)*(1/4*f2^2) = (598/2.17*10^-3)*(1/4(353.23)^2)

=55.21 cm

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