Consider a harmonic oscillator with mass=0.1kg, k=50N/m , h-bar=1.055x10-34
Let this oscillator have the same energy as a mass on a spring, with the same k and m, released from rest at a displacement of 5.00 cm from equilibrium. What is the quantum number n of the state of the harmonic oscillator?
The Attempt at a Solution
ok, so i was wondering, why can't i use angular momentum quantization?
meaning mvr = n(h-bar)
so i divide both sides by r2
i get m(k/m)-1/2=n(h-bar)/r2
make n subject, sub in all values, i get 5.29*1031.. which happens to be twice the correct answer.
so did i make a careless mistake somewhere? or was i never been able to use this formula? thanks!