Question

A string of linear mass density 2.19 g/m is stretched by the weight of an adjustable mass m as shown. Near the end of the string a vibrator is attached at a constant but unknown frequency; the length of the string which vibrates is 2.39 m. For some values of the mass m n, the string resonates with the vibrator

Answer 1

f = N / (2 L) * (T / (m/l))^1/2       where T is tension and N the number of 1/2 wavelengths
f = N / (2 L) * k (T)^1/2      where k is a constant since m/l is a constant
f2 / f1 = (N - 1) / N * (T2 / T1)^1/2     dividing equations
f2 / f1 = (N - 1) / N * (M2 / M1)^1/2  since the tension is proportional to M
N / (N - 1) = f1 / f2 * (1.21 / 1)^1/2 = 1.11 f1 / f2    rearranging
N = 1.11 N - 1      since frequency is constant
N = 10     the number of 1/2 wavelengths
y (wavelength) = 2.39 / 5 = .48 m
So for 1 kg number of wavelengths is 5 with 11 antinodes
(2 antinodes at each end and 9 between the ends)
f = N / (2 L) * (T / (m/l))^1/2 = 10 / 4.78 * (9.8 / .00219)^1/2 = 140 Hz