3
$\begingroup$

My book says that functions which represent a wave motions satisfy the differential equation :

$$ \frac{\partial^2 y} {\partial^2t} = v^2 \frac{\partial^2 y} {\partial^2x} $$

where $v$ is the velocity of the wave.

But I don't understand its physical reason.

As a side question, I just wanted some intuition for the formula for the speed of the wave on a string i.e. : $ v = \sqrt{ T/m}$ Like why should it hold true, its physical reason.

$\endgroup$

1 Answer 1

3
$\begingroup$

Changing coordinates $u= x+vt$ $w= x-vt$ the written equation transforms to $$\frac{\partial^2 y }{\partial u \partial w}=0\tag{1}\:.$$ Assuming that $(t,x)$ ranges in the whole $\mathbb R^2$ that is equivalent to say that $(u,w)$ ranges in the same set, the solutions of (1) can easily showed to be all the function of the form $$y(u,w) = f(u) + g(w)\:.$$ Coming back to the initial variables we have: $$y(t,x)= f(x+ vt) + g(x-vt)\:.$$ The physical meaning of that solution is evident. Varying $t$, the first function $f(x+vt)$ describes a rigid translation, from the right to the left, of the profile $y=f(x)$, with constant velocity $v>0$. Similarly, varying $t$, the second function $g(x-vt)$ describes a rigid translation from the left to the right of the profile $y=g(x)$, with constant velocity $v>0$.

Concerning your last question, a careful analysis just based on "F=ma" applied to a small portion of a string with linear mass density $m$ and tension $T$, establishes that, for small deformations along the vertical axis (in comparison with the deformation along the horizontal axis), the profile of the string evolves with the mentioned equation.

ADDENDUM. I have some spare time before my lecture, so I can complete the last part of my answer. Consider the string described by $y=y(x)$ and a small portion of string of endpoints $x$ and $x+d$. If the vertical slope is small with respect to $d$, the mass of the portion is approximatively $md$, where $m$ is the constant linear density mass of the string. Along the vertical direction, the equation of motion of the portion is, assuming the tension $T$ remaining constant in absolute value (but with different directions at different points): $$m d\frac{\partial^2 y}{\partial t^2}|_{x'} = T \frac{\partial y}{\partial x}|_{x+d} - T \frac{\partial y}{\partial x}|_{x}\:.\tag{1}$$ Above $x'$ is the middle point of the portion, $\frac{\partial y}{\partial x} = \tan \alpha(x)$ is the slope of the string at the generic point $x$. The net force on the portion is due to the tension vector acting at the two endpoints and pulling them towards opposite directions with different slopes. The projection of the tension vector along $y$ (that vector is always tangent to the string) should be $T \sin \alpha(x)$, however, for small slopes $\tan \alpha = \sin \alpha$ and the identity $$T \sin \alpha(x) = T \frac{\partial y}{\partial x}$$ is approximatively true. If $d$ is small, (1) can be rewritten as $$m d\frac{\partial^2 y}{\partial t^2}|_x = T \frac{\partial^2 y}{\partial x^2}|_xd$$ that is $$\frac{\partial^2 y}{\partial t^2} = \frac{T}{m} \frac{\partial^2 y}{\partial x^2}\:.$$ This is the wanted equation.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.