Asked by Tim
A string is fixed at both ends and vibrating at 120 Hz, which is its third harmonic frequency. The linear density of the string is 4.9x10-3 kg/m, and it is under a tension of 3.6 N. Determine the length of the string.
Answers
Answered by
ButtholeDr
This question seems to be missing the radius of the string.
From the equation v = sqrt(T/mu)
we can solve for mu to get: mu = T/V^2
We already have tension, to find velocity we use v = lambda*frequency
Because its a third harmonic frequency the lambda(wavelength) is 2/3 of the string's length.
V= 2/3L(120Hz)
V= 80L
mu = 3.6 N/80L
mu represents mass per unit length
mu = mass/Length
We don't have mass so we substitute density*volume
to get: mu =d*pi*r^2*L/ L
Now we plug this back in to our first equation to solve for L.
d*pi*r^2*L/ L=3.6 N/80L
L=.045/(4.9x10-3*pi*r^2)
From the equation v = sqrt(T/mu)
we can solve for mu to get: mu = T/V^2
We already have tension, to find velocity we use v = lambda*frequency
Because its a third harmonic frequency the lambda(wavelength) is 2/3 of the string's length.
V= 2/3L(120Hz)
V= 80L
mu = 3.6 N/80L
mu represents mass per unit length
mu = mass/Length
We don't have mass so we substitute density*volume
to get: mu =d*pi*r^2*L/ L
Now we plug this back in to our first equation to solve for L.
d*pi*r^2*L/ L=3.6 N/80L
L=.045/(4.9x10-3*pi*r^2)
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