The string has an original length of L, mass density of μ, and is under a tension of T which gives it a fundamental frequency of f1. The string is then folded at its midpoint, halving its length and doubling its mass density. With the tension still at T, the question is to determine the new fundamental frequency f1'.

1. A string has the length of L and the linear mass density ofμ. Suppose that under a certain
tension T, this string has the fundamental frequency f1. Then the string is folded at the midpoint
and twisted tightly, so that both the length and the linear mass density are changed. The tension
of the string is controlled to stay the same. Suppose that the fundamental frequency is f1
’ now,
solve for f1
’.
Solution:
v = √(T/μ)
For fundamental frequency,λ= 2L
So f1 = v/λ= (1/2L)[√(T/μ)]
Since the string is folded at the midpoint, the length becomes one half of the original:
L’ = L/2.
Since μ= m/L, m does not change but L is reduced to one half,
μ’= m/(L/2) = 2(m/L) = 2μ.
So f1
’ = (1/2L’) [√(T/μ’)]
= [1/2(L/2)][√(T/2μ)]
= (1/L)[√(T/2μ)]
= 2(1/2L)[√(1/2)][√(T/μ)]
= 2[√(1/2)]·{(1/2L)[√(T/μ)]}
=√(2)·f1
Answer: f1
’=√(2)·f1 = 1.41f1