Question

How to to find wave speed of a string using derivation of second newton law ?

Answer 1

Let us first draw an image showing a part of a string with forces.

7 2 dm 2

Let us apply Newton's second law in the y-direction

mdy

The sum of the forces in the y-direction is

$F_y=T\ sin\theta_2-T\ sin\theta_1$

Now, from small angle approximation

oy sin10-tan

$\Rightarrow F_y=T\left (\frac{\partial y}{\partial x} \right )_2-T\left (\frac{\partial y}{\partial x} \right )_1$

Let us assume the mass per unit length of the string is

$\mu=\frac{m}{L}$

So, the mass of the string element is

$dm=\mu dx$

So, the acceleration in the y-direction is the rate of change of velocity in the y-direction

$a_y=\frac{\partial v_y}{\partial t}=\frac{\partial^2y}{\partial t^2}$

so, we can write Newton's second law in the y-direction as

Oy ду Oy CT

Upon rearranging

$\frac{\partial^2y}{\partial t^2}=\frac{T}{\mu}\frac{\left (\frac{\partial y}{\partial x} \right )_2-\left (\frac{\partial y}{\partial x} \right )_1}{dx}$

Now we have been using the subscript 1 to identify the position x, and 2 to identify the position (x+dx). So the numerator in the last term on the right is the difference between the (first) derivatives at these two points. When we divide it by dx, we get the rate of change of the first derivative with respect to x, which is, by definition, the second derivative, so we have derived the wave equation:

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

The solution to this wave equation is

y -A sin (k-t)

The partial derivatives are

o-y 0t2 --w2A sin (kz-wt)

02 k'A sin (kr- wt)

Which gives us

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

In traveling waves the wave velocity is defined by

$v=\frac{\omega}{k}$

Which gives us the velocity of the wave as

$v=\frac{\omega}{k}=\sqrt\frac{T}{\mu}$