∂2y∂x2+k2y=0the solution for this will be:y = Asin(kx - wt) + Bcos(kx - wt)∂y∂t=-Aω(cos[kx-ωt])+Bω(sin[kx-ωt])=&gt;&nbsp;∂2y∂t2=-Aω2(sin[kx-ωt])-Bω2(cos[kx+ωt])=&gt;&nbsp;∂2y∂t2=-ω2[y]=&gt;&nbsp;1c2∂2y∂t2=-ω2c2yalso, c = w/ktherefore,&nbsp;k2=w2c2=&gt;&nbsp;1c2∂2y∂t2=-k2y=&gt;&nbsp;1c2∂2y∂t2=∂2y∂x2[since y is a solution for x].therefore, displacement of the standing wave expression satisfies the time independent form.

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2022-03-28

Answer & Explanation

user_27qwe

Skilled2022-07-02Added 375 answers

$\frac{\partial^{2} y}{\partial x^{2}} + k^{2} y = 0$

the solution for this will be:

y = Asin(kx - wt) + Bcos(kx - wt)

$\frac{\partial y}{\partial t} = - A ω (\cos [k x - ω t]) + B ω (\sin [k x - ω t])$

=> $\frac{\partial^{2} y}{\partial t^{2}} = - A ω^{2} (\sin [k x - ω t]) - B ω^{2} (\cos [k x + ω t])$

=> $\frac{\partial^{2} y}{\partial t^{2}} = - ω^{2} [y]$

=> $\frac{1}{c^{2}} \frac{\partial^{2} y}{\partial t^{2}} = - \frac{ω^{2}}{c^{2}} y$

also, c = w/k

therefore, $k^{2} = \frac{w^{2}}{c^{2}}$

=> $\frac{1}{c^{2}} \frac{\partial^{2} y}{\partial t^{2}} = - k^{2} y$

=> $\frac{1}{c^{2}} \frac{\partial^{2} y}{\partial t^{2}} = \frac{\partial^{2} y}{\partial x^{2}}$

[since y is a solution for x].

therefore, displacement of the standing wave expression satisfies the time independent form.

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