Answer:f1 = 7.90Hz, f2= 15.811Hz, f3 = 23.71Hz.
Explanation: length of string = 10m, mass of string = 100g = 0.1kg, T= 250N
We need the velocity of sound wave in a string when plucked with a tension T, this is given below as
v = √T/u
Where u = mass /length = 0.1/ 10 = 0.01kg/m
Hence v = √250/0.01, v = √25,000 = 158.11
a) at the lowest frequency.
At the lowest frequency, the length of string is related to the wavelength with the formulae below
L = λ/2, λ= 2L.
λ = 2 * 10
λ = 20m.
But v = fλ where v = 158.11m/s and λ= 20m
f = v/ λ
f = 158.11/ 20
f = 7.90Hz.
b) at the first frequency.
The length of string and wavelength for this case is
L = λ.
Hence λ = 10m
v = 158.11m/s
v= fλ
f = v/λ
f = 158.11/10
f = 15.811Hz
c) at third frequency
The length of string is related to the wavelength of sound with the formulae below
L =3λ/2, hence λ = 2L /3
λ = 2 * 10 / 3
λ = 20/3
λ= 6.67m
v = fλ where v = 158.11m/s, λ= 6.67m
f = v/λ
f = 158.11/6.67
f = 23.71Hz.
