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.

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

To find the lowest frequency that permits standing waves in both strings with a node at the junction, you can follow these steps:

Step 1: Find the velocities in each string:
The velocity of a wave is given by v = √(T/μ), where T is the tension and μ is the linear density of the string.

For string 1:
T1 = 190.0 N
μ1 = (m/L)1 = 6.00x10^-2 kg/m

v1 = √(T1/μ1) = √(190.0/6.00x10^-2) m/s

For string 2:
T2 = 190.0 N
μ2 = (m/L)2 = 1.50x10^-2 kg/m

v2 = √(T2/μ2) = √(190.0/1.50x10^-2) m/s

Step 2: Find the ratio between n1 and n2:
Use the equation (n1v1)/2L1 = (n2v2)/2L2, where n is the number of loops and L is the length of the string.

Rewriting the equation, we have:
(n1v1)/L1 = (n2v2)/L2

Since we need to find the lowest frequency, we can assume n1 = 1 (the fundamental frequency) and find the corresponding value of n2.

(n1v1)/L1 = (n2v2)/L2

(1*v1)/L1 = (n2v2)/L2

Simplifying further, we have:
v1/v2 = (n2L1)/(L2)

Rearranging, we get:
n2 = (v1/v2) * (L2/L1)

Step 3: Substitute the given values and calculate the ratio:
Substitute the values of v1, v2, L1, and L2 into the equation found in Step 2 and solve for n2.

n2 = (v1/v2) * (L2/L1)
n2 = (√(190.0/6.00x10^-2) / √(190.0/1.50x10^-2)) * (1.25 / 3.75)

Simplify the expression and calculate n2.

This will give you the ratio between n1 and n2.

To find the lowest frequency that permits standing waves in both strings with a node at the junction, you need to consider the fundamental frequency (n=1) for each string. The fundamental frequency corresponds to the standing wave pattern with the fewest number of loops.

First, let's find the velocity (v) in each string. The velocity of a wave can be determined using the equation v = √(T/μ), where T is the tension and μ is the linear density of the string.

For the first string:
T1 = 190.0 N
μ1 = 6.0 x 10^-2 kg/m

Using the equation, v1 = √(T1/μ1) = √(190.0/6.0 x 10^-2) = √(3166.67) ≈ 56.29 m/s

For the second string:
T2 = 190.0 N
μ2 = 1.5 x 10^-2 kg/m

Using the equation, v2 = √(T2/μ2) = √(190.0/1.5 x 10^-2) = √(12666.67) ≈ 112.62 m/s

Now that we have the velocities v1 and v2, we can find the ratio between n1 and n2.

The equation for the frequency of a standing wave (f) can be written as f = nv/2L, where n is the number of loops and L is the length of the string.

For the first string:
f1 = (n1v1)/2L1, where n1 is the number of loops in the first string.
Since we are looking for the fundamental frequency (n=1), we can rewrite the equation as f1 = v1/2L1.

For the second string:
f2 = (n2v2)/2L2, where n2 is the number of loops in the second string.
Again, for the fundamental frequency (n=1), we can rewrite the equation as f2 = v2/2L2.

To find the ratio between n1 and n2, simplify the equations:
f1 = v1/2L1
f2 = v2/2L2

Divide the two equations:
f1/f2 = (v1/2L1)/(v2/2L2)

The velocity and length terms cancel out:
f1/f2 = (v1/v2) x (2L2/2L1)
f1/f2 = (v1/v2) x (L2/L1)

Substitute the given values:
f1/f2 = (56.29/112.62) x (1.25/3.75)
f1/f2 = 0.5 x 0.3333
f1/f2 ≈ 0.1667

So the ratio between n1 and n2 is approximately 0.1667.