Question

Problem 1: A particle of mass 5 10-8 kg is attached to a massless spring whose spring constant is 2 10-3N/m. Maximum displacement of the particle from the equilibrium position is 3 - 10-9m. The world is one- dimensional. (a) Find the (classical) amplitude of these oscillations. Yes, funny question, but I had to ask (b) Calculate the quantum of energy ho of this harmonic oscillator. he oea rthe prtic estimale the cormesponding uanitum umher. (d) If this harmonic oscillator were cooled all the way to its ground state, what would be the spatial extent of its wave function?

Answer 1

Given k = 2 times 10^-3 N/m m = 5 times 10^-8 kg x = 3 times 10^-9 m Acceleration of the block. a = F - kx/m = variable at a = 0 x = F/k Rightarrow F = kx = 2 time axis 10^-3 times 3 times 10^-9 = 6 times 10^-12 N (a) a = v dv/dx Rightarrow v dv/dx = F - kx/m v dv = F - kx/m dx \r\n integral v dv = integral F - kx/m dx at maximum elongation (amplitude) velocity will be zero. Integral_0^v v dv = integral_0^A F - k_x/m dx A = Amplitude. 0 = 1/m [Fx - kx^2/2]_0^A FA - kA^2/2 = 0 FA = kA^2/2 Rightarrow A = 2F/k A = 2 times 6 times 10^-12/2 times 10^-3 A = 6 times 10^-9 m (b) since w = squareroot k/m = squareroot 2 times 10^-3/5 times 10^-8 = squareroot 0.4 times 10^5 squareroot 4 times 10^4 = 2 times 10^2 per sec