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July 24, 2014

July 24, 2014

Posted by **Tina** on Monday, July 11, 2011 at 6:49pm.

3.75 m 1.25m

_________________=====================

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.

- Physics -
**Damon**, Monday, July 11, 2011 at 7:36pmSince 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

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