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An electron experiences a quasi-elastic force `kx` and `a` 'friction force' `yx` in the field of electromagnetic radiation. The `E`component of the field varies as `E = E_(0) cos omegat`. Neglecting the action of the magnetic component of the field, find, (a) the motion equation of the electron, (b) the mean power absored by the electron, the frequency at which that power is maximum and the expression for the maximum mean power. |
Answer» The equation of the electron can (under the stated conditions) be written as `mddot(x) + gamma dot(x) + kx = eE_(0) cos omega t` To solve this equation we shell find it convenient to use complex displacements. Consider the equation `mddot(z) + gamma dot(z) + kz = eE_(0) e^(-i omegat)` `z = (eE_(0)e^(-i omegat))/(-m omega^(2) - i gamma omega + k)` (we igone transients) Writing `beta = (gamma)/(2m), omega_(0)^(2) = (k)/(m)` we find `z = (eE_(0))/(m)e^(-i omegat)// (omega_(0)^(2) - omega^(2) - 2i beta omega)` Now `x =`1 Real part of `z` `= (eE_(0))/(m). (cos (omegat + varphi))/(sqrt((omega_(0)^(2) - omega^(2))^(2) + 4beta^(2) omega^(2))) = a cos (omega t + varphi)` where `tan varphi = (2 beta omega)/(omega^(2) - omega_(0)^(2))` `(sin varphi =- (2 beta omega)/(sqrt((omega_(0)^(2) - omega_(0)^(2))^(2) + 4beta^(2) omega^(2))))` (b) We calculate the power absorded as `P = lt Edot(x) gt = lt eE_(0)cos omegat (-omega a sin (omegat + varphi)) gt` `=eE_(0).(eE_(0))/(m)(1)/(2). (2 beta omega)/((omega_(0)^(2) - omega^(2))^(2) + 4beta^(2) omega^(2)).omega = ((eE_(0))/(m))^(2) (beta m omega^(2))/((omega_(0)^(2) - omega^(2))^(2) + 4beta^(2) omega^(2))` This is clearly maximum when `omega_(0) = omega` because `P` can be written as `P = ((eE_(0))/(m))^(2) (beta omega)/((omega_(0)^(2)/ omega^(2)-omega)^(2) + 4beta^(2))` and `P_(max) = (m)/(4 beta) ((eE_(0))/(m))^(2)` for `omega = omega_(0)`. `P` can also be calculated from `P = lt gamma dot(x). dot(x) gt` `= (gamma omega^(2) a^(2)//2) = (beta m omega^(2) (eE_(0)//m)^(2))/((omega_(0)^(2) - omega^(2))^(2) + 4beta^(2)omega^(2))`. |
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