Question

A string with mass per unit length of 0.003 kg/m is plucked with an amplitude of 2 cm. The string is 30 cm long, and under a tension of 30 newtons. What are the frequencies of the first 5 harmonics?

Answer 1

Answer:

First harmonics = 333.33N

Second harmonics = 500.01N

Third harmonics = 666.68N

Fourth harmonics = 833.35N

Fifth harmonics = 1000.02N

Explanation:

The formula for calculating the fundamental frequency in string is expressed as;

$F_0 = \frac{1}{2L}\sqrt{\frac{T}{m} }$ where;

L is the length of the string = 30cm = 0.3m

T is the tension in the string = 30N

m is the mass per unit length of the string = 0.003kg/m

Get the fundamental frequency first by substituting the given values into the formula;

$F_0 = \frac{1}{2(0.3)}\sqrt{\frac{30}{0.003} }\\F_0 = \frac{1}{0.6}\sqrt{10,000}}\\F_0 = \frac{1}{0.6} * 100$

F0 = 166.67N

Harmonics are the integral multiples of the fundamental frequency.

First harmonics F1 = 2F0 = 2(166.67) = 333.33N

Second harmonics F2 = 3F0 = 3(166.67) = 500.01N

Third harmonics F3 = 4F0 = 4(166.67) = 666.68N

Fourth harmonics F4 = 5f0 = 5(166.67) = 833.35N

Fifth harmonics F5 = 6f0 = 6(166.67) = 1000.02N