A string with a mass/length of 0.00200 kg/m is stretched between two points with a tension of 115 N.If the distance between the points is changed by 1.0%,then what is the percentage change of the frequency?
A) 0.50%
B) 0.75%
C) 1.00%
D) 1.25%
E) 1.50%

Question

Quizplus · Accepted Answer

The frequency of a stretched string is given by:
f = (1/2L)√(T/μ)
where L is the length of the string, T is the tension and μ is the mass per unit length.

Now, if the tension is kept constant and the length is changed by ΔL, then the new frequency f' is given by:

f' = (1/2(L+ΔL))√(T/μ)

So, the percentage change in frequency is given by:

Δf/f = (f' - f)/f = [(1/2(L+ΔL))√(T/μ) - (1/2L)√(T/μ)]/[(1/2L)√(T/μ)]

= (ΔL/L)/(2√(T/μ)) = 0.01/(2√(115/0.002)) = 0.375%

Therefore, the answer is C, 1.00%.

A String with a Mass/length of 0