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Apr 8, 2015 at 18:34 comment added user36790 @Valter Moretti: Sir, I'll be grateful to you if you please explain to me how the wave-equation is analogous to the second law. And sir, 2nd derivative of position means acceleration; what do these 2nd derivatives mean?
Mar 30, 2015 at 14:42 comment added Valter Moretti Use $\mu \frac{\partial^2y}{\partial t^2} = T_0 \frac{\partial^2 y}{\partial x^2}$ in the second term in the right-hand side and take advantage of $\frac{\partial \frac{\partial y}{\partial x}}{\partial t}= \frac{\partial^2 y}{\partial x \partial t}$
Mar 30, 2015 at 13:46 comment added mwa1 Actually I don't easily see how you got to equation (3). I'm stuck at $\frac{\partial e(x,t)}{\partial t} = T_0 \frac{\partial y}{\partial x} \frac{\partial \left(\frac{\partial y}{\partial x}\right)}{\partial t} + \mu \frac{\partial y}{\partial t}\frac{\partial ^2 y}{\partial t^2}$
Mar 30, 2015 at 12:46 vote accept mwa1
Mar 30, 2015 at 11:52 history edited Valter Moretti CC BY-SA 3.0
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Mar 30, 2015 at 11:45 history edited Valter Moretti CC BY-SA 3.0
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Mar 30, 2015 at 11:38 history answered Valter Moretti CC BY-SA 3.0