Answer to Question #148864 in Mechanics | Relativity for Gaurav

Question

Derive the speed of transverse waves on a uniform string, with the help of suitable diagram.

Answer 1

Explanations & Calculations

• For an any transverse wave of the form $\small$ $\small y(x,t)=f(x \pm vt)$, the derivative form can be written as $\frac{\partial^2y}{\partial x^2}\small(x,t)=\frac{1}{v^2} \frac{\partial^2y}{\partial^2 t^2}\small(x,t) \cdots (1)$

• Consider an elemental segment (AB) of a uniform thread taut to a tension of T and total mass of M. (First figure)
• Mass of the elemental segment is $\small \Delta M = \frac{M}{L}\times \Delta x$
• The quantity $\small \frac{M}{L}$ is known as the linear density m
• Then consider it after plucked & has entered into a wavy motion. (second figure) Here the elemental segment is located in a slanted position between AB.
• Then the thread experiences a net tension ($\small T_{net}$ ) which comprises of the previous tension and the restoring force which acts to reverse the change made in to the system. And it has the usual horizontal & vertical components.
• Since the segment do not travel horizontally, those vertical components remains equal & cancels each other.
• And for the vertical components apply Newton's second law upwards

$\qquad\qquad \begin{aligned} \small F_1 -F_2 &= \small \Delta m a=\Delta m \frac{\partial^2 y}{\partial t^2}\\ \small F_1 -F_2 &= \small m \Delta x \frac{\partial^2 y}{\partial t^2 }\cdots (1) \end{aligned}$

• From the wave form above,

$\qquad\qquad \begin{aligned} \small \tan\theta _1 =\small \frac{F_1}{F_x} \end{aligned}$ and $\qquad\qquad \begin{aligned} \small \tan\theta_2 =\frac{F_2 }{F_x} \end{aligned}$

• These tan values are the gradients of the waveform at those given points. Therefore,

$\qquad\qquad \begin{aligned} \small \tan \theta _1 &= \small \frac{\partial y}{\partial x}\Big|_B \end{aligned}$ and $\qquad\qquad \begin{aligned} \small \tan\theta_2 =\small\frac{\partial y}{\partial x}\Big|_A \end{aligned}$

• Substituting these in (1),

$\qquad\qquad \begin{aligned} \small F_x\Bigg[\frac{\partial y}{\partial x}\Big|_B-\frac{\partial y}{\partial x}\Big|_A\Bigg]&= \small m\Delta x\frac{\partial^2y}{\partial t^2}\\ \small \frac{\Bigg[\frac{\partial y}{\partial x}\Big|_B-\frac{\partial y}{\partial x}\Big|_A\Bigg]}{\Delta x}&= \small \frac{m}{F_x}\frac{\partial^2y}{\partial t^2} \end{aligned}$

• And taking limit of this equation as $\small \lim_{\Delta x\to0}$ ,

$\qquad\qquad \begin{aligned} \small \frac{\partial^2y}{\partial x^2 } &= \small \frac{m}{F_x}\frac{\partial ^2 y}{\partial t^2 }\cdots(2) \end{aligned}$

• Comparing (2) with (1),

$\qquad\qquad \begin{aligned} \small \frac{m}{F_x} &= \small \frac{1}{v^2}\\ \small |v| = \sqrt{\frac{F_x}{m}} \end{aligned}$

• This horizontal component is the tension of the string ,therefore,

$\qquad\qquad \begin{aligned} \small |v| = \sqrt{\frac{T}{m}} \end{aligned}$