Answer:
Explanation:
The fundamental frequency of the string f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.
[tex]V = \sqrt{\frac{T}{\mu} }[/tex] where T is the tension in the string and [tex]\mu[/tex] is the density of the string
Given T = 600N and [tex]\mu[/tex] = 0.015 g/cm = 0.0015kg/m
[tex]V = \sqrt{\frac{600}{0.0015} }\\ \\V = \sqrt{400,000}\\ \\V = 632.46m/s[/tex]
The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.
L = 2m
Substituting the derived values into the formula f₀ = V/2L
f₀ = 632.46/2(2)
f₀ = 632.46/4
f₀ = 158.12 Hertz
Hence the fundamental frequency of the string is 158.12 Hertz
