GATE EC 2022 Set 1 — Question 30

The given wave equation is:$$\frac{\partial^2 f(x, t)}{\partial t^2} = 10000 \frac{\partial^2 f(x, t)}{\partial x^2}$$This is a one-dimensional wave equation of the form $\frac{\partial^2 f}{\partial t^2} = c^2 \frac{\partial^2 f}{\partial x^2}$, where $c$ is the wave speed. Comparing the given equation, we have $c^2 = 10000$, so $c = 100$.The general solution to the one-dimensional wave equation is given by D'Alembert's formula: $f(x,t) = g(x-ct) + h(x+ct)$, where $g$ and $h$ are arbitrary twice-differentiable functions. For our equation, $c=100$, so the solutions must be of the form $f(x,t) = g(x-100t) + h(x+100t)$.Let's check each option:Option (A): $f(x,t) = e^{-(x-100t)^2} + e^{-(x+100t)^2}$
This function is a sum of two terms: $g(x-100t) = e^{-(x-100t)^2}$ and $h(x+100t) = e^{-(x+100t)^2}$. Both terms are of the correct form $g(x-ct)$ and $h(x+ct)$ with $c=100$. Since the wave equation is linear, a sum of solutions is also a solution. Thus, this is a solution.
To verify explicitly for a term $f_1(x,t) = g(x-ct)$: Let $u = x-ct$.
$\frac{\partial f_1}{\partial t} = g'(u) (-c)$ and $\frac{\partial^2 f_1}{\partial t^2} = g''(u) (-c)^2 = c^2 g''(u)$.
$\frac{\partial f_1}{\partial x} = g'(u) (1)$ and $\frac{\partial^2 f_1}{\partial x^2} = g''(u) (1)^2 = g''(u)$.
So, $\frac{\partial^2 f_1}{\partial t^2} = c^2 \frac{\partial^2 f_1}{\partial x^2}$ is satisfied. The same applies to $h(x+ct)$. Therefore, (A) is a solution.Option (B): $f(x,t) = e^{-(x-100t)} + 0.5e^{-(x+1000t)}$
The first term $e^{-(x-100t)}$ is of the form $g(x-100t)$ and is a solution. However, the second term $0.5e^{-(x+1000t)}$ is of the form $h(x+c't)$ where $c' = 1000$. Since $c' \neq c$, this term does not satisfy the wave equation with $c=100$. Therefore, the sum is not a solution. This statement is incorrect.Option (C): $f(x,t) = e^{-(x-100t)} + \sin(x + 100t)$
The first term $e^{-(x-100t)}$ is of the form $g(x-100t)$ and is a solution (as shown in option A's verification). The second term $\sin(x+100t)$ is of the form $h(x+100t)$. Let's verify for $f_2(x,t) = \sin(x+ct)$:
$\frac{\partial f_2}{\partial t} = c \cos(x+ct)$ and $\frac{\partial^2 f_2}{\partial t^2} = -c^2 \sin(x+ct)$.
$\frac{\partial f_2}{\partial x} = \cos(x+ct)$ and $\frac{\partial^2 f_2}{\partial x^2} = -\sin(x+ct)$.
So, $\frac{\partial^2 f_2}{\partial t^2} = c^2 \frac{\partial^2 f_2}{\partial x^2}$ is satisfied. Since both terms are solutions, their sum is also a solution. This statement is correct.Option (D): $f(x,t) = e^{j100\pi(-100x+t)} + e^{j100\pi(100x+t)}$
Let's analyze the first term: $f_1(x,t) = e^{j100\pi(t-100x)}$. This is of the form $g(t-100x)$.
If $f(x,t) = g(t-cx)$, then $\frac{\partial^2 f}{\partial t^2} = g''(t-cx)$ and $\frac{\partial^2 f}{\partial x^2} = c^2 g''(t-cx)$.
Substituting these into the wave equation $\frac{\partial^2 f}{\partial t^2} = c^2 \frac{\partial^2 f}{\partial x^2}$ gives $g''(t-cx) = c^2 (c^2 g''(t-cx))$, which implies $1 = c^4$. This is only true if $c=1$, but here $c=100$. Therefore, $g(t-cx)$ is not a solution to $\frac{\partial^2 f}{\partial t^2} = c^2 \frac{\partial^2 f}{\partial x^2}$ unless $c=1$. The first term is not a solution.
Alternatively, for a plane wave $e^{j(\omega t - kx)}$ to be a solution, we need $\omega = ck$. For the first term, $\omega = 100\pi$ and $k = 100\pi \times 100 = 10000\pi$. So, $c = \omega/k = (100\pi)/(10000\pi) = 1/100$. This is not $c=100$. Thus, the first term is not a solution. Since one term is not a solution, the sum is not a solution. This statement is incorrect.The final answer is $\boxed{A, C}$