Since there is a node at the junction the two strings might just as well be strung beside each other or in different rooms. The only thing the same about them is the tension of 190 N.
string 1 L =3.75 so 2L = 7.5 and mu = 6*10^-2
f1 = (1/2L)sqrt(F/mu)
= (1/(3.75*2) ) sqrt (190/6*10^-2)
=2.373 Hz
string 2 L = 1.25 so 2L - 2.5 and mu = 1.5*10^-2
f1 = (1/2.5)sqrt(190/1.5*10^-2)
= 14.236 Hz
We need the least common multiple of those two frequencies.
In fact 14.236/2.373 = 5.999
which I call six
so I claim that 14.236 Hz is it
with n = 1 for string 1 and n = 6 for string 2
Question: Two strings at different lengths and linear densities are joined together. They are stretched so that the tension in each string is 190.0 N. The free ends are fixed in place. Find the lowest frequency that permits standing waves in both strings with a node at the junction. The standing wave pattern in each string may have a different number of loops.
3.75 m 1.25m
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6*10^-2 kg/m 1.5*10^-2 kg/m
(this is kinda what the picture looks like if it helps but it seems really confusing on the computer)
Equations I should use (there may be others):
f=nv/2L
(n1v1)/2L1=(n2v2)/2L2
Other information I am given:
L1=3.75
L2=1.25
n1=?
n2=?
v1=?
v2=?
(m/L)=6.00x10^-2 kg/m
(m/L)2=1.50x10^-2 kg/m
I need to create a ratio between n1 and n2.
I don’t understand how I get v and then how to find a ratio between n1 and n2.
1 answer
1 answer