GATE 2022 EC Q.30: solution and answer

GATE 2022 · ECQ.30 · 1 mark · MSQ

Consider the following wave equation,

\frac{\partial ^{2} f ( x , t )}{\partial t ^{2}} = 10000 \frac{\partial ^{2} f ( x , t )}{\partial x ^{2}}

Which of the given options is/are solution(s) to the given wave equation?

(A)
$f (x, t) = e^{- (x - 100 t)^{2}} + e^{- (x + 100 t)^{2}}$
(B)
$f (x, t) = e^{- (x - 100 t)} + 0.5 e^{- (x + 1000 t)}$
(C)
$f (x, t) = e^{- (x - 100 t)} + sin (x + 100 t)$
(D)
$f (x, t) = e^{j 100 π (- 100 x + t)} + e^{j 100 π (100 x + t)}$

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Answer and solution

ANSWER

Options (A) and (C) are solutions of $f_{tt} = 10000 f_{xx}$ . Every term in them depends only on $x - 100 t$ or $x + 100 t$ , which is a wave travelling at $c = 10000 = 100$ .

Official answer: A, CSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Compare with the standard 1-D wave equation $f_{tt} = c^{2} f_{xx}$ . Here $c^{2} = 10000$ , so the wave speed is $c = 100$ .
Define the residual $R [f] = f_{tt} - 10000 f_{xx}$ . $f$ is a solution exactly when $R [f] \equiv 0$ , and $R$ is linear: $R [f_{1} + f_{2}] = R [f_{1}] + R [f_{2}]$ . Now take a travelling-wave term $ϕ (x + v t)$ , where $v$ may be negative. Then $ϕ_{tt} = v^{2} ϕ^{''}$ and $ϕ_{xx} = ϕ^{''}$ , so $R [ϕ (x + v t)] = (v^{2} - 10000) ϕ^{''} .$ If $ϕ^{''}$ is not identically zero, the term is a solution only when $v = \pm 100$ , i.e. when it is a function of $x - 100 t$ or of $x + 100 t$ . This is d'Alembert's form $f = g (x - 100 t) + h (x + 100 t)$ .
(A): Both terms use $ϕ (u) = e^{- u^{2}}$ , one with $v = - 100$ and one with $v = + 100$ . Each has $R = 0$ , so by linearity the sum has $R = 0$ . Solution.
(B): The first term, $e^{- (x - 100 t)}$ , has $v = - 100$ , so its residual is $0$ . The second term has $v = 1000$ , so its residual is $R [0.5 e^{- (x + 1000 t)}] = (1 0^{6} - 1 0^{4}) \cdot 0.5 e^{- (x + 1000 t)} = 4.95 \times 1 0^{5} e^{- (x + 1000 t)} .$ The total residual is $0 + 4.95 \times 1 0^{5} e^{- (x + 1000 t)}$ , which is never zero. Not a solution.
(C): $e^{- (x - 100 t)}$ has $v = - 100$ and $sin (x + 100 t)$ has $v = + 100$ . Each residual is $0$ , so the sum has $R = 0$ . Solution.
(D): Write each term as $e^{j (ω t + k x)}$ with $ω = 100 π$ . The first term has $k = - 1 0^{4} π$ and the second has $k = + 1 0^{4} π$ . For such a term, $f_{tt} = - ω^{2} f$ and $f_{xx} = - k^{2} f$ , so $R = (1 0^{4} k^{2} - ω^{2}) f = (1 0^{12} π^{2} - 1 0^{4} π^{2}) f = 1 0^{4} π^{2} (1 0^{8} - 1) f .$ Both terms have the same factor, so the residual of the whole option is $R = 1 0^{4} π^{2} (1 0^{8} - 1) [e^{j (100 π t - 1 0^{4} π x)} + e^{j (100 π t + 1 0^{4} π x)}] = 2 \cdot 1 0^{4} π^{2} (1 0^{8} - 1) e^{j 100 π t} cos (1 0^{4} π x) .$ This is not identically zero; at $x = 0$ , $t = 0$ it equals $2 \cdot 1 0^{4} π^{2} (1 0^{8} - 1)$ . The two terms therefore do not cancel each other. Not a solution. The reason: $- 100 x + t = - 100 (x - \frac{t}{100})$ , so these waves travel at $\frac{ω}{∣ k ∣} = \frac{1}{100}$ , not $100$ .
The correct options are (A) and (C).

KEY IDEA

Every solution of $f_{tt} = c^{2} f_{xx}$ has the form $g (x - c t) + h (x + c t)$ , with $c = 100$ here. To test a term, write its argument as a multiple of $x + v t$ and check that $v = \pm 100$ . If a term fails, work out the residual of the whole sum: the bad terms' residuals could in principle cancel, so check that they do not.

WATCH OUT FOR

In (D), taking the 100 that multiplies $x$ as the wave speed. The argument is $- 100 x + t = - 100 (x - t /100)$ , so the speed is $1/100$ . The mistake runs the other way in (B): reading $x + 1000 t$ as close enough to $x + 100 t$ when its speed is $1000$ . A third error is to reject a sum only because each of its terms fails. Terms that fail can still cancel, so always compute the residual of the whole expression.

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GATE 2022 EC Q.30: solution and answer

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