learning object

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