linear wave equation\n- derive linear wave equation by using wave function, y(x, t)=a sin(kx + ωt)\nconsider…
linear wave equation\n- derive linear wave equation by using wave function, y(x, t)=a sin(kx + ωt)\nconsider acceleration of a particle,\na = ∂²y(x,t)/∂t² = ∂/∂t u = ∂/∂t (∂y/∂t)\n- aω² sin(kx + ωt)\n∂²y/∂x² / ∂²y/∂t² = k²/ω² = 1/v² ⇒ ∂²y/∂x² = 1/v² ∂²y/∂t²\nconsider slope of the wave function, slope = ∂y(x,t)/∂x = ak cos(kx+ωt)\ncurvature = ∂²y(x,t)/∂x² = -ak² sin(kx+ωt)\n∂²y/∂t² = v² ∂²y/∂x²
Answer
Explanation:
Step1: Find the second - order time - derivative
The wave function is $y(x,t)=A\sin(kx + \omega t)$. First, find the first - order time - derivative $\frac{\partial y(x,t)}{\partial t}=A\omega\cos(kx+\omega t)$. Then, find the second - order time - derivative $\frac{\partial^{2}y(x,t)}{\partial t^{2}}=-A\omega^{2}\sin(kx+\omega t)$.
Step2: Find the second - order space - derivative
Find the first - order space - derivative $\frac{\partial y(x,t)}{\partial x}=Ak\cos(kx+\omega t)$. Then, find the second - order space - derivative $\frac{\partial^{2}y(x,t)}{\partial x^{2}}=-Ak^{2}\sin(kx+\omega t)$.
Step3: Relate the derivatives
We know that the wave speed $v = \frac{\omega}{k}$, or $\omega=vk$. Substitute $\omega = vk$ into the second - order time - derivative $\frac{\partial^{2}y(x,t)}{\partial t^{2}}=-A(vk)^{2}\sin(kx+\omega t)$. Since $\frac{\partial^{2}y(x,t)}{\partial x^{2}}=-Ak^{2}\sin(kx+\omega t)$, we can get $\frac{\partial^{2}y(x,t)}{\partial t^{2}}=v^{2}\frac{\partial^{2}y(x,t)}{\partial x^{2}}$, which is the linear wave equation.
Answer:
The linear wave equation $\frac{\partial^{2}y(x,t)}{\partial t^{2}}=v^{2}\frac{\partial^{2}y(x,t)}{\partial x^{2}}$ is derived as above.