The fundamental frequency on a vibrating string is given by:
[tex]f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}[/tex]
where
L is the length of the string
T is the tension
[tex]\mu[/tex] is the mass per unit length of the string
Keeping this equation in mind, we can now answer the various parts of the question:
(a) The fundamental frequency will halve
In this case, the length of the string is doubled:
L' = 2L
Substituting into the expression of the fundamental frequency, we find the new frequency:
[tex]f'=\frac{1}{2(2L)}\sqrt{\frac{T}{\mu}}=\frac{1}{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{2}[/tex]
So, the fundamental frequency will halve.
(b) the fundamental frequency will decrease by a factor [tex]\sqrt{2}[/tex]
In this case, the mass per unit length is doubled:
[tex]\mu'=2\mu[/tex]
Substituting into the expression of the fundamental frequency, we find the new frequency:
[tex]f'=\frac{1}{2L}\sqrt{\frac{T}{2 \mu}}=\frac{1}{\sqrt{2}}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{\sqrt{2}}[/tex]
So, the fundamental frequency will decrease by a factor [tex]\sqrt{2}[/tex].
(c) the fundamental frequency will increase by a factor [tex]\sqrt{2}[/tex]
In this case, the tension is doubled:
[tex]T'=2T[/tex]
Substituting into the expression of the fundamental frequency, we find the new frequency:
[tex]f'=\frac{1}{2L}\sqrt{\frac{2T}{\mu}}=\sqrt{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\sqrt{2}f[/tex]
So, the fundamental frequency will increase by a factor [tex]\sqrt{2}[/tex].
