motion of the piano string

ismail

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Sep 28, 2015
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Hi,

Could you please help me for b and c?



. . . . .\(\displaystyle \dfrac{\partial^2 y}{\partial x^2}\, =\, \dfrac{1}{c^2}\, \dfrac{\partial^2 y}{\partial t^2}\)

subject to the boundary condition \(\displaystyle \, y(0,\, t)\,=\, y(\ell,\, t)\, =\, 0.\,\) (The boundary condition implies that the string is fixed at both ends.) The initial conditions are different, however.

a) Solve for the motion of the harpsichord string, assuming that the initial conditions are

. . . . .\(\displaystyle y(x,\, 0)\, =\, 1\, -\, 2\, \dfrac{\left|\strut x\, -\, \dfrac{\ell}{2}\right|}{\ell}\). . . . .\(\displaystyle \mbox{ for }\, 0\, \leq\, x\, \leq \ell\)

. . . . .\(\displaystyle \dfrac{\partial y(x,\, t)}{\partial t}\Bigg\rvert_{t\, =\, 0}\, =\, 0\). . . . . . . . . . .\(\displaystyle \mbox{ for }\, 0\, \leq\, x\, \leq \ell\)

b) Solve for the motion of the piano string, assuming that the initial conditions are

. . . . .\(\displaystyle y(x,\, 0)\, = \, 0\). . . . . . . . . . . . . . . . . . .\(\displaystyle \mbox{ for }\, 0\, \leq\, x\, \leq \ell\)

. . . . .\(\displaystyle \dfrac{\partial y(x,\, t)}{\partial t}\Bigg\rvert_{t\, =\, 0}\, =\, 1\, -\, 2\, \dfrac{\left|\strut x\, -\, \dfrac{\ell}{2}\right|}{\ell}\). . . . .\(\displaystyle \mbox{ for }\, 0\, \leq\, x\, \leq \ell\)

c) Explain why a harpsichord sounds different from a piano. Which has a more pure sound? (That is, which has more energy in its fundamental mode relative to the other modes?)



Thank you.
 
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