Beats from Different String Lengths

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SUMMARY

The discussion centers on calculating the beat frequency produced by two strings of different lengths, specifically when one string is 0.57 cm longer than the other. The fundamental frequency of the shorter string is given as 226 Hz, and the wave speed is 34.9 m/sec. The correct approach involves using the formula for frequency, f1 = v/2L, to determine the lengths of both strings and subsequently calculate the beat frequency using fbeat = |f1 - f2|. The error identified in the calculations was due to incorrectly subtracting the length difference instead of adding it to find the longer string's length.

PREREQUISITES
  • Understanding of wave propagation and frequency calculations
  • Familiarity with the formula f = v/2L for fundamental frequency
  • Knowledge of beat frequency calculation using fbeat = |f1 - f2|
  • Basic unit conversions between centimeters and meters
NEXT STEPS
  • Review the principles of wave mechanics and string vibrations
  • Practice calculations involving different string lengths and their frequencies
  • Explore the effects of varying wave speeds on frequency and beat frequency
  • Investigate real-world applications of beat frequencies in musical instruments
USEFUL FOR

Students studying physics, particularly in wave mechanics, musicians interested in understanding sound frequencies, and educators teaching concepts related to waves and vibrations.

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Homework Statement



Two strings which are fixed at both ends are identical except that one is 0.57 cm longer than the other. Waves on both of these string propagate with a speed of 34.9 m/sec and the fundamental frequency of the shorter string is 226 Hz.

a) What is frequency of the beat that would result if these two strings were plucked at the same time?

c) What is the beat frequency if the length difference is now 0.72cm?

Homework Equations



f1 =v/2L
fn= n*(v/2L)
fbeat= abs f 1- f 2

The Attempt at a Solution


So I tried using the first equation with f1 being 226Hz and then used the velocity given to solve for the first length. Which I found to be 0.0772123m. Then I subtracted .57cm from that to get length 2 (0.0829123) and used the same equation again except it will now be f2= 2* (34.9/2*0.0829123). This gave me the second freqeuncy of 420.9262362. When I subtract this from 226, it gives me the wrong answer for f beat.
Please advise me as I am not sure at what I am doing wrong at this point. Thanks in advance
 
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Hi rpp293!
Welcome to PF! :smile:


rpp293 said:
Then I subtracted .57cm from that to get length 2 (0.0829123) and used the same equation again...


You have to add (not subtract) 0.0057 m to 0.077 m to get the length of the longer string.
 

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