Question

What is the role of tension in wave motion.

Answer 1

Answer on Question #38028, Physics, Mechanics

Question:

What is the role of tension in wave motion?

Answer:

Tension determines a phase speed of waves. It can be shown by the following procedure. Let us consider a string with linear density of $\mu$ . We can write a Newton's second law in the vertical direction as

$$F _ {y} = m a _ {y} = T \sin \theta_ {1} - T \sin \theta_ {2},$$

where $m$ is a mass of small piece of string

and $T$ is a tension force. If one replaces each sine by derivative due to smallness of angles for the case of linear oscillations, we get the following identity:

$$\mu d x \frac {\partial^ {2} y}{\partial t ^ {2}} = T \left(\left. \frac {\partial y}{\partial x} \right| _ {x = x _ {2}} - \left. \frac {\partial y}{\partial x} \right| _ {x = x _ {1}}\right),$$

or

$$\frac {\partial^ {2} y}{\partial t ^ {2}} = \frac {T}{\mu} \left. \frac {1}{d x} \left(\frac {\partial y}{\partial x} \right| _ {x = x _ {2}} - \frac {\partial y}{\partial x} \right| _ {x = x _ {1}}$$

The last fraction can be replaced by the second derivative and finally we obtain the wave equation

$$\frac {\partial^ {2} y}{\partial t ^ {2}} = \frac {T}{\mu} \frac {\partial^ {2} y}{\partial x ^ {2}}.$$

Solution of this equation can be found as $y = A\sin (kx - \omega t)$ . Substituting it one obtains the following relation

$$\frac {T}{\mu} = \left(\frac {\omega}{k}\right) ^ {2}$$

Because $\frac{\omega}{k} = v_{phase}$ by definition,

$$v_{phase} = \sqrt{\frac{T}{\mu}}.$$

Thus, one can see that the phase speed of a wave is determined by a restoring property (tension force $T$) and inertial property (mass density $\mu$).