# Standing waves question

1. Apr 27, 2014

### Just_some_guy

This is my first post on the forum, and I hope you guys can help me

My questions is this,
There are two strings which are connected with constant tension everywhere, but they have different linear mass densities. The second mass density is double that of the first (μ1= 2μ1). A wave propagates along the "lighter" string with an amplitude of 10 cm and a wavelength of 10 cm. I have to find the wavelength of the wave propagated through the second string, assuming frequency is the same everywhere, and also what is the amplitude of the transmitted wAve?

Relevant equations which I have used are

v= ω/k = λω/2π where k= 2π/λ

Also v^2 = T/μ

All I have done so far is to notice that if the second mass density is double the first then the second velocity is a factor of sqrt2/2 smaller than the first? So would λ then simply be the same factor larger than that of the first wavelength as frequency is constant?

If this is the case then my answer was 14 cm

I haven't yet begun the second part as I wanted to do the first part first

Cheers!

2. Apr 27, 2014

### Just_some_guy

I hope my question is clear

Last edited: Apr 27, 2014
3. Apr 27, 2014

### Just_some_guy

Ok I have now calculated a value for lambda 2

I got 14 cm

Using v1/λ1= v2/λ2

And put v1= sqrt2 / 2 and v2= 1 because that's the factor between the two as tension is uniform and frequency is equal over all parts

Last edited: Apr 27, 2014
4. Apr 27, 2014

### dauto

I think your ratio between the speeds is upside down. Since the wave passes from the lighter spring to the heavier one, the wave slows down and shrinks by a factor of √2. Besides this minor mistake, the procedure you used is correct. The second question is harder to solve.

5. Apr 27, 2014

### Just_some_guy

So my new wavelength would then be

7 cm rather than 14, by getting rid of my factor of 1/2

Do you have any suggestions for the second part I'm not so sure?:(

6. Apr 27, 2014

### Just_some_guy

Would the answer not have to be 14 cm

As v= fλ then if we have velocity decreasing then to compensate and keep frequency the same then wavelength must surely increase? Not decrease also?

7. Apr 27, 2014

### dauto

You're confused. If the left side of the equation decreases than the right side must ALSO DECREASE to compensate.

8. Apr 27, 2014

### Just_some_guy

So 7 cm would be the correct answer

9. Apr 27, 2014

### dauto

So it seems.

10. Apr 28, 2014

### Just_some_guy

That makes sense, I was mixing up the formula I was using:(
Anyways thanks, I also found I good way to find amplitude.....

A2/A1 = 2k1/ k1+ k2

And calculated each term according to

k= 2π/λ