Mechanical Waves & Wave Equation

[verd]

Hi, I'm having a little bit of difficulty understanding exactly what to do to get to an answer in section a of this problem. It asks to show that the given function satisfies the wave equation... I have the wave equation. How do I go about 'showing' that it satisfies the wave equation?

Do I just differentiate it twice? ...If so, to which respect to I differentiate it to?


Here's the problem:

You want to measure the mass m of an object, but you don’t have a scale. You therefore decide to attach the object to a string of mass ms and length L, as in the figure, and to generate standing waves on the string (pay attention to the orientation of the axes on the figure!). The wave function that describes the standing wave is given by:
$$y(x,t)=\cos(\frac{2\pi}{\lambda}x+\phi)\cos(2\pi ft)$$


where λ is the wavelength, f is the frequency, and φ is the phase.


a) Show that the wave function y(x,t) satisfies the
wave equation and from it derive and expression
for the speed of propagation of the wave in terms of
the given quantities.

 

[nrqed]

Hi, I'm having a little bit of difficulty understanding exactly what to do to get to an answer in section a of this problem. It asks to show that the given function satisfies the wave equation... I have the wave equation. How do I go about 'showing' that it satisfies the wave equation?

Do I just differentiate it twice? ...If so, to which respect to I differentiate it to?


Here's the problem:

You want to measure the mass m of an object, but you don’t have a scale. You therefore decide to attach the object to a string of mass ms and length L, as in the figure, and to generate standing waves on the string (pay attention to the orientation of the axes on the figure!). The wave function that describes the standing wave is given by:
$$y(x,t)=\cos(\frac{2\pi}{\lambda}x+\phi)\cos(2\pi ft)$$


where λ is the wavelength, f is the frequency, and φ is the phase.


a) Show that the wave function y(x,t) satisfies the
wave equation and from it derive and expression
for the speed of propagation of the wave in terms of
the given quantities.

calculate ${\partial^2 y(x,t) \over \partial t^2}$ and ${\partial^2 y(x,t) \over \partial x^2}$. The ratio of the first over the second will give you the square of the speed (which will obviously be $\lambda^2 f^2$).

 

[kyle8921]

To show that the wave function y(x,t) satisfies the wave equation, we need to differentiate it twice with respect to both x and t. This will give us the second derivative of the function, which is necessary for the wave equation.

First, let's differentiate with respect to x:

\frac{\partial^2 y}{\partial x^2} = \frac{\partial}{\partial x} (-\frac{2\pi}{\lambda} \sin(\frac{2\pi}{\lambda}x+\phi)\cos(2\pi ft)) = (-\frac{2\pi}{\lambda})^2 \cos(\frac{2\pi}{\lambda}x+\phi)\cos(2\pi ft)

Now, let's differentiate with respect to t:

\frac{\partial^2 y}{\partial t^2} = \frac{\partial}{\partial t} (-\frac{2\pi}{\lambda} \cos(\frac{2\pi}{\lambda}x+\phi)\sin(2\pi ft)) = (-2\pi f)^2 \cos(\frac{2\pi}{\lambda}x+\phi)\cos(2\pi ft)

We can see that both of these derivatives are equal to the original function y(x,t) multiplied by a constant. This is exactly what the wave equation states: the second derivative of the function with respect to both x and t is equal to the function multiplied by a constant (in this case, (-\frac{2\pi}{\lambda})^2 and (-2\pi f)^2).

To derive the expression for the speed of propagation, we can use the wave equation in its simplest form:

\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}

where v is the speed of propagation. We can rearrange this equation to solve for v:

v = \frac{1}{\sqrt{\frac{\partial^2 y}{\partial x^2}} \frac{\partial^2 y}{\partial t^2}}

Substituting in the derivatives we found earlier, we get:

v = \frac{1}{\sqrt{(-\frac{2\pi}{\lambda})^2 \cos(\frac{2\pi}{\lambda}x+\phi