String wave and small angle approximation?

[genxium]

Homework Statement

I'm so confused about the derivation of the famous equation $$v=\sqrt{\frac{T}{u}}$$, I tried to derive it by myself but failed, then I turned to wikipedia but the derivation there really gave me a shock!

http://en.wikipedia.org/wiki/Vibrating_string"

I have no idea why it can assume that's available to use small angle approximation in the figure, if the very small segment is not chosen from the top of the wave, angle $$\alpha$$ and $$\beta$$ should not be taken as small angles.

Homework Equations

T: Tension in the string, u: Linear density of the string mass

The Attempt at a Solution

I tried to make another assumption: For each small segment $$\Delta l$$ in the string, 1st , it doesn't move along the x-axis so the acceleration on x-axis for each point is always 0, 2nd , the tension in the string is not equal everywhere. But just as everyone knows, this assumption doesn't lead to a meaningful answer.

 

[PeterO]

Homework Statement

I'm so confused about the derivation of the famous equation $$v=\sqrt{\frac{T}{u}}$$, I tried to derive it by myself but failed, then I turned to wikipedia but the derivation there really gave me a shock!

http://en.wikipedia.org/wiki/Vibrating_string"

I have no idea why it can assume that's available to use small angle approximation in the figure, if the very small segment is not chosen from the top of the wave, angle $$\alpha$$ and $$\beta$$ should not be taken as small angles.

Homework Equations

T: Tension in the string, u: Linear density of the string mass

The Attempt at a Solution

I tried to make another assumption: For each small segment $$\Delta l$$ in the string, 1st , it doesn't move along the x-axis so the acceleration on x-axis for each point is always 0, 2nd , the tension in the string is not equal everywhere. But just as everyone knows, this assumption doesn't lead to a meaningful answer.

The reason it is reasonable to assume small angles [as in the highlighted section above] is because the amplitude is small - the crests are tiny.

Consider a guitar string" half a wavelength is the length of the string, and the vibrating string does not even get close to the adjacent string - indeed max amplitude is seldom more than a mm or two.

 

[genxium]

The reason it is reasonable to assume small angles [as in the highlighted section above] is because the amplitude is small - the crests are tiny.

Consider a guitar string" half a wavelength is the length of the string, and the vibrating string does not even get close to the adjacent string - indeed max amplitude is seldom more than a mm or two.

Thanks a lot!