Question

Can you do (b) and (c) only thank you

PrOBleM: SoLuTiONS To THE WAvE EQuATION a) By direct substitution determine which of the following functions satisfy the wave equation 1. g(z, t)-A cos(kr - wt) where A, k, w are positive constants 2. h(z,t)-Ae-(kz-wt)2 where A, k, ω are positive constants 3. p(x, t) A sinh(kx-wt) where A, k,w are positive constants 4. q(z, t) - Ae(atut) where A,a, w are positive constants 5. An arbitrary function: f(x, t) - f(kx -wt) where k and w are positive constants. (Hint: Be careful with your derivative rules.) b) What similarities among functions that satisfy the wave equation do you observe? Do you no- tice any differences between functions that satisfy the wave equation and those that do not? c) Verify by direct substitution that an arbitrary linear combination of two functions that satisfy the wave equation will also satisfy the wave equation. That is, if f(z, t) and g(x,t) both satisfy the wave equation you are to show that the function F(x, t)-αυζ,t)+Ag(x, t) also satisfies the wave equation provided a and ß are constants

Answer 1

b)The functions that satisfy the wave equation all have the form $y=f(x-vt)$ i.e. the variables x and t occur only in the combination (x-vt).This accounts for similarities and differences between functions that satisfy the wave equation and those that do not.

c) The wave equation is given by:

$\frac{\partial^2 y}{\partial x^2}=1/v^{2}\frac{\partial^2y }{\partial t^2}$.$y_{1}=f(x,t)$ and $y_{2}=g(x,t)$ are solutions of the wave equation.

Then $y=\alpha y_{1}+\beta y_{2}$ is also a solution of the above equation as is proved below by substitution.

Now we have

$\frac{\partial^2 (\alpha y_{1}+\beta y_{2})}{\partial x^2}=\alpha \frac{\partial^2 y_{1}}{\partial x^2}+\beta \frac{\partial^2 y_{2}}{\partial x^2}=\alpha *1/v^{2}\frac{\partial^2 y_{1}}{\partial x^2}+\beta *1/v^{2}\frac{\partial^2 y_{2}}{\partial x^2}=1/v^{2}(\frac{\partial^2 \alpha y_{1}}{\partial x^2})+1/v^{2}\frac{\partial^2 \beta y_{2}}{\partial x^2}=1/v^{2}(\frac{\partial^2 (\alpha y_{1}+\beta y_{2})}{\partial t^2})=1/v^{2}\frac{\partial^2 y}{\partial t^2}$.

i.e. $\frac{\partial^2 y}{\partial x^2}=1/v^{2}\frac{\partial^2 y}{\partial t^2}$ which is the required proof.