Two strings have different lengths and linear densities, as the drawing shows. T

Question

Two strings have different lengths and linear densities, as the drawing shows. They are joined together and stretched so that the tension in each string is 239 N. The free ends of the joined string are fixed in place. Find the lowest frequency that permits standing waves in both strings with a node at the junction. The standing wave pattern in each string may have a different number of loops.

                                                               node

--------------------------------------------------- / .................................................
3.75m long                                                     1.25m long
6*10^-2 kg/m                                                 1.5*10^-2 kg/m

Explanation / Answer

Linear densities: µ1 = 6*10^-2 Kg/m µ2 = 1.5*10^-2 Kg/m Tension in each string T = 239 N Let n and n' are number of loops in the left and right strings respectively. Now we can write (n/l1) Sqrt[T/µ1] = (n'/l2) Sqrt[T/µ2] n/n' =(l1/l2) Sqrt[µ1 / µ2] = 3 Sqrt[6/1.5] = 6/1 So n = 6 and n' = 1. The frequency of the left string which is equal to the fundamental frequency of the right string is f = n (1/2l) Sqrt[T/µ1] = 6 (1/2*3.75) Sqrt[239/0.06] = 50.49 Hz.

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