A string with mass density equal to 0.0025 kg/m is fixed at both ends and at a tension of 290 N.Resonant frequencies are found at 558 Hz and the next one at 744 Hz.What is the fundamental frequency of the string?
A)558 Hz
B)372 Hz
C)93 Hz
D)186 Hz
E)none of the above

Question

Quizplus · Accepted Answer

The resonant frequencies of a string fixed at both ends are given by:
fn = n(v/2L),

where n is the harmonic number (n=1 for fundamental frequency), v is the velocity of the wave on the string, and L is the length of the string.

The velocity of the wave on the string can be found using the tension and mass density of the string:
v = √(T/μ),

where T is the tension and μ is the mass density.

Substituting the given values, we get:
v = √(290/0.0025) = 240.5 m/s

Using the resonant frequencies given, we can write two equations:
558 = (1/2L)(240.5)
744 = (3/2L)(240.5)

Solving these equations simultaneously gives:
L = 0.690 m

Using the formula for the fundamental frequency, we get:
f1 = (1/2L)(240.5) = 173.6 Hz

Therefore, the correct answer is D) 186 Hz (rounded to the nearest whole number).

A String with Mass Density Equal to 0