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saturnsalien
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This isn't actually a homework or coursework problem, but the style of the question is similar so I'm posting it here. Anyways, here goes. Consider a string of length L clamped at both ends, with one end at x=0 and the other at x=L. The displacement of the oscillating string can be described by the following equation:
\frac{\partial^2 \Psi}{\partial t^2}=\frac{\partial^2 \psi}{\partial x^2}
<br /> \textrm{Given that at t=0:}\\<br /> \>\>\>\> \psi(x,0)=\frac{2xh}{L},0\leq x\leq \frac{L}{2}\\<br /> \>\>\>\> \psi(x,0)=\frac{2xh}{L},(L-x),\frac{L}{2}\leq x\leq L\\<br /> \textrm{Show:}\\<br /> \>\>\>\> \psi(x,t)=\sum_{m=1}^\infty\sin\left(\frac{m\pi x}{L}\right)\cdot\cos\omega_mt\cdot\left(\frac{8h}{\pi^2m^2}\right)\cdot\sin\left(\frac{\pi m}{2}\right)<br />
So, how do we go about doing that?
\frac{\partial^2 \Psi}{\partial t^2}=\frac{\partial^2 \psi}{\partial x^2}
<br /> \textrm{Given that at t=0:}\\<br /> \>\>\>\> \psi(x,0)=\frac{2xh}{L},0\leq x\leq \frac{L}{2}\\<br /> \>\>\>\> \psi(x,0)=\frac{2xh}{L},(L-x),\frac{L}{2}\leq x\leq L\\<br /> \textrm{Show:}\\<br /> \>\>\>\> \psi(x,t)=\sum_{m=1}^\infty\sin\left(\frac{m\pi x}{L}\right)\cdot\cos\omega_mt\cdot\left(\frac{8h}{\pi^2m^2}\right)\cdot\sin\left(\frac{\pi m}{2}\right)<br />
So, how do we go about doing that?
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