Chem 113A HW2 Solutions

Chem 113A HW2 Solutions - Chem 113A Homework Assignment...

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Chem 113A - Homework Assignment Week 2 Solutions 1) a. This is a linear, homogeneous differential equation so we guess the solution: ?(?) = ? 𝑟? Plugging into our differential equation we obtain: ?? 2 + 𝑘 = 0 ? = ±𝑖√ 𝑘 ? = 0 This gives the general solution: ?(?) = ? 1 ? 𝑖√ 𝑘 ? ? + ? 2 ? −𝑖√ 𝑘 ? ? Employing the boundary conditions yields: ?(0) = ? 1 + ? 2 = 0 ? 1 = −? 2 ?? ?? (0) = ? 1 𝑖√ 𝑘 ? − ? 2 𝑖√ 𝑘 ? = ? 0 ? 1 = ? 0 2 ? 𝑘 And our solution is ?(?) = ? 0 𝑚 𝑘 ? 𝑖√ 𝑘 𝑚 𝑡 −? 0 𝑚 𝑘 ? −𝑖√ 𝑘 𝑚 𝑡 2 = ? 0 ? 𝑘 sin(√ 𝑘 ? ?) b. The max displacement is ? 0 ? 𝑘 and the frequency is 1 2𝜋 𝑘 ? . c. ?(0) = 𝛿 and ?? ?? (0) = 0 d. ?(?) = 𝛿 cos (√ 𝑘 ? ?) The maximum displacement of the oscillator is 𝛿 , which makes sense since at this displacement the spring’s total energy is the potential energy (no initial velocity). Thus it cannot gain energy and wind up at a larger displacement. The frequency is 𝑘 ? and it does not depend on the initial displacement or initial velocity. e. In this idealized case it does but if we had to take into account any dampening forces like friction or air resistance the model would fail. The ball also can’t displace equivalently in both directions since the it would collide with the wall on the left.

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2) Travelling wave: 𝛹(?, ?) = ??𝑖?(𝑘? − ??) Standing wave: 𝛹(?, ?) = ??𝑖?(𝑘?) cos(??) = ?
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