Finding Frequencies in Two Strings Stuck Together

[Hypnos_16]

Homework Statement

Two strings have different lengths (L1 = 3.920 m and L2 = 1.960 m) and linear densities (mL1 = 2.64E-1 kg/m and mL2 = 6.60E-2 kg/m), as the drawing below shows.

|........|
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|----------------------------------|
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(Okay, only the "-"s are the line, it's pretty much just two strings that are stuck together, with different linear densities)

They are joined together and stretched so that the tension in each string is 192 N. The free ends of the joined string 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.

L1 = 3.92 m
L2 = 1.96 m
mL1 = 0.264 kg/m
mL2 = 0.0660 kg/m
T = 192 N

Homework Equations

v = √(T / µ)
v = ƒ (Landa)

The Attempt at a Solution

v = √(T / µ)
v = √(192 / 0.264)
v = 27.0 m

v = √(T / µ)
v = √(192 / 0.0660)
v = 54.0 m

I found the speed in both strings using their proper linear density. After getting two speeds and having two different wavelengths resulting in two completely different frequencies. Then i realized perhaps the speed should be the same, since they're attached. But even then the frequency isn't the same.

Help?

 

[anathema]

Thank you for sharing your problem with us. It seems like you are on the right track with your calculations. However, there are a few things that you need to consider in order to find the correct frequency for standing waves in both strings.

Firstly, it is important to note that when two strings are joined together, the tension in the string is not evenly distributed. In other words, the tension in the string with a higher linear density will be greater than the tension in the string with a lower linear density. This means that the speed of the waves in each string will be different.

To find the correct frequency, you will need to use the speed of the waves in the string with the higher linear density. This is because the standing wave pattern will have a shorter wavelength in this string, and therefore a higher frequency.

So, in your calculations, you should use the speed of 54.0 m/s, which you found for the string with the higher linear density of 0.0660 kg/m. This will give you a frequency of 13.5 Hz.

I hope this helps you in finding the correct solution. Keep up the good work and don't hesitate to ask for help if needed.
Scientist