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1

I was just researching this kind of questions, since the derivation found in most textbooks, in terms of tension, seems a little unrelated to material properties. Three things: As pointed out, the tension T needs to be inside the square root The velocity of sound in the string material is unrelated to the (phase) velocity of the wave. As the formula shows, ...

2

Of course there can be (and are) traveling waves on a string - when you pluck anywhere except at the center, you are generating an asymmetrical impulse that will travel up and down the string. The fact is that your "pluck" consists of many different frequencies, and as these travel back and forth two things happen: the highest frequencies will be damped ...

1

They are traveling waves for a very short period of time until they reach at the ends of the string and then reflect.Then they produce a standing wave.A standing wave consists of tavelling waves.

4

The equation you quote is an approximation that is only valid if the horizontal force on the string remains the same. In practice that is not the case - and your concern is valid. The increase in length $\Delta \ell =\ell(\frac{1}{\cos\theta}-1)$; how much additional force that generates depends on the unstretched length of the string (or equivalently on ...

0

Firstly, tension force itself depends on the elastic/rigid property of the string. So, the properties are implicit in the mathematical representation of the tension. Let a small part of the string be acted by tension $T_0$ from both sides of the string & hence it remains in static equilibrium as in the first picture. Now, when that part is stretched, ...

3

D'Alembert equation reads, for the considered case, $$\mu\frac{\partial^2 y}{\partial t^2} = T_0 \frac{\partial^2 y}{\partial x^2}\:.\tag{1}$$ It is nothing but $F=ma$ along the vertical direction ($y$). Here $y$ denotes the small deformation of the string along the vertical direction from the stationary (horizontal) configuration. We disregard horizontal ...

0

I think in this example, you have to find the tension "in terms of". To do so, you have to do it using the static equilibrium equation which gives you the solution that you wrote. So, here it works the other way around. You first find the tension in terms of $T_0$ and THEN you find the new length of the wire. Why do you have to do it this way? Well, you ...

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