Two identical piano wires have a fundamental frequency of 600 Hz when kept under the same tension.What fractional increase .....

Question

Two identical piano wires have a fundamental frequency
of 600 Hz when kept under the same tension.What fractional
increase in the tension of one wire will lead to the occurrence
of 6.0 beats/s when both wires oscillate simultaneously?

beaukamrath · Answer

The beat frequency: ∆f = 6 Hz 



 If we increase string tension, then, the frequency of the sound on the 

 string will be higher. thus, f' = f + ∆f = 600 + 6 = 606 Hz 



 The speed of a traveling sound wave on a string is given by: 



 v = √T/µ 



 With the tension on the 2nd string increased, v' = √T'/µ 



 where, tension T' = T + ∆T. 



If we take the ratio of wave speeds on the strings: 



 v' / v = √T'/µ / √T/µ = √T'/T 



Squaring both sides: (v'/v)² = (T + ∆T) / T = 1 + ∆T/T 



∆T/T = (v' / v)² - 1 



But, the velocity is directly proportional to the frequency, since, v = f λ. 



So that without loss of generality, we can rewrite the tension ratio above as: 



∆T/T = (f' / f)² - 1 = (606/600)² - 1 = 1.0201 - 1 = 0.0201 



∆T/T = 2.01%

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Two identical piano wires have a fundamental frequency of 600 Hz when kept under the same tension.What fractional increase .....

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