Question

$If \: y = A \: sin(ωt - kx), \: then \: \\ find \: the \: value \: of \: \frac{ \frac{ {d}^{2}y }{d {t}^{2} } }{ \frac{ {d}^{2}y }{d {x}^{2} } }$

#Answer the question with steps#

Answer 1

We are given an equation:

$y = A\sin (\omega t -kx)$

This is the equation of a traveling wave. y denotes the transverse displacement of the particles, x denotes the position of a specific particle on the wave, and t is time. 

Also, A is the Amplitude of the wave. $\omega$ and k are other constants. 

In simple words, the equation gives the displacement y of a particle located at position x at time t. 

Here, both x and t are variables. So we will be using Partial Differentiation.
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Some Info on Partial Differentiation


Suppose we have a function:
$z=x^2y^3$

where both x and y are variables. We cannot simply find $\frac{dz}{dx}$ or $\frac{dz}{dy}$, because somehow the variables are dependent on each other. 

So, we use Partial Differentiation. We assume one variable as constant and differentiate with respect to the other variable. 

Consider again:

$z=x^2y^3$ 

If we want to find the partial derivative of z w.r.t x then we will write:

$\frac{\partial z}{\partial x} = 2xy^3$ 

Similarly, if we want to find partial derivative of z w.r.t. y , then we assume x as constant, and so:

$\frac{\partial z}{\partial y} = x^2 (3y^2) = 3x^2y^2$


The symbol $\partial$ can be read as "del" or "partial dee". There are quite some other ways to read it though. [People at my place call it "del" ]
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Coming to question. Our equation is:

$y=A\sin (\omega t-kx)$

To write $\frac{dy}{dx}$ isn't exactly wrong, but isn't also exactly right. We will instead use the symbol for partial derivatives.

The symbol $\frac{\partial y}{\partial t}$ represents that while differentiating, we are treating x as constant. This means we are considering a particle at a fixed position. We are thus analyzing a specific particle over time.


Similarly, $\frac{\partial y}{\partial x}$ represents that while differentiating, we must treat t as constant. That is, we are considering different particles of the wave at a fixed time.


Also the concept we are going to use is:

$\text{If } y = a \sin (bx+c) \\ \\ \implies \frac{dy}{dx} = a\cos (bx+c) \times \frac{d}{dx}(bx+c) = ab\cos(bx+c) \\ \\ \text{Similarly } \frac{d^2y}{dx^2} = -ab^2 \sin (bx+c)$



Here, we have:


$y=A\sin (\omega t-kx) \\ \\ \implies \frac{\partial y}{\partial t} = A\omega \cos (\omega t-kx) \\ \\ \implies \frac{\partial^2 y}{\partial t^2} = -A\omega^2\sin (\omega t -kx)$


Similarly, we have:


$y=A\sin (\omega t-kx) \\ \\ \implies \frac{\partial y}{\partial x} = -Ak \cos (\omega t-kx) \\ \\ \implies \frac{\partial^2 y}{\partial x^2} = -Ak^2\sin (\omega t -kx)$


Finally, our answer would be:

$\displaystyle Ans = \frac{\frac{\partial^2y}{\partial t^2}}{\frac{\partial^2y}{\partial x^2}} \\ \\ \\ \implies Ans = \frac{-A\omega^2\sin (\omega t -kx)}{-Ak^2\sin (\omega t -kx)} \\ \\ \\ \implies \boxed{Ans = \frac{\omega^2}{k^2}}$

Actual Answer ends here.

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Physical Meaning: [Extra Info]


$\rightarrow \omega$ is known as Angular Frequency. It is related to Time Period of a wave, T, as follows:

$\omega = \frac{2\pi}{T}$


$\rightarrow k$ is known as the Angular Wave Number. It is related to the wavelength of the wave, $\lambda$ as follows:

$k = \frac{2\pi}{\lambda}$

When we write $\frac{\omega}{k}$, we are actually writing:

$\frac{\omega}{k} = \frac{\frac{2\pi}{T}}{\frac{2\pi}{\lambda}} = \frac{\lambda}{T}$

This is nothing but Wave Velocity. That is, the wave takes a time equal to $T$ to travel a distance $\lambda$. 

If we write wave velocity as $v$, we can write:

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

So, our answer in the above question actually is:

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

For the sake of our question, however, we would just stick to $\frac{\omega^2}{k^2}$.