Mandolin Strings
Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 591 Hz. The peg o
Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 591 Hz. The peg of one string slips slightly, and the tension in it drops to 194 N. How Many beats per second are heard?
The speed of a wave on a string goes like the square root of the Tension in the string.
The wavelength is constant (because it is defined by the length of the string which isn’t changing).
For all waves, speed = frequency * wavelength
Therefore, the frequency goes like the square root of tension. So the new frequency is given by:
f2 = f1 * sqrt (T2/T1)
where f1 and T1 are the original frequency and tension, and T2 is the new tension.
The beat frequency is just the difference between the two freqencies.
f1 – f2 = f1 (1 – sqrt (T2/T1))
If you Taylor expand the square roots and throw away terms higher than first order (okay in this case because the tension doesn’t change too much), you get the simpler formula given in the answer below.
Vivaldi / I Solisti Veneti: Concerto in G Major for Two Mandolins, Strings and Continuo, P. 133
