Question

A distance of 8.00 cm is measured between two adjacent nodes of a standing wave on a 32.0 cm long string.

a) In which harmonic number n is the string vibrating?

b0 Find the frequency (In Hz) of this harmonic if the string has a mass of 2.05 x 10^-2 kg and a tension of 855 N. (give answer in Hz)

Answer 1

Nooll Node CX 8cm NI L 32cm Distance between two adjacent nodes is equal to half of the wavelength = 8cm Since the length of the string is 32 cm which is 4 times of 8cm , the distance between two nodes therefore there are 4 loops in the string String is vibrating at 4 times of the fundamental frequency therefore string is vibrating in fourth harmonic b. Wave speed is given by V= Here T the tension and u is the linear mass density which is given by

Mass u = Length M L u = So V= U TL V - M A = 8cm 2 2 = 16cm 2 = 0.16m frequency of this harmonic V f= - = X TL M 2 2 1 f= (855) x (0.32) 2.05 x 10-2 0.16 f= 722.04Hz