Question

A string is stretched to a length of 361 cm and both ends are fixed. If the density of the string is 0.039 g/cm, and its tension is 923 N, what is the fundamental frequency? Answer in units of Hz.

Answer 1

Answer:

67.38 Hz

Explanation:

Using,

F = 1/(2L)[√(T/ρ)].............. Equation 1

Where F = fundamental frequency of the string, L= length of the string, T = Tension in the string, ρ = density of the string.

Given: L = 361 cm = 3.61 m, ρ = 0.039 g/cm = 0.0039 kg/m, T = 923 N.

Substitute into equation 1

F = 1/(2×3.61)[√(923/0.0039)]

F = 1/7.22(√236666.67)

F = 1/7.22(486.48)

F = 67.38 Hz

Hence the fundamental frequency of the string = 67.38 Hz