Question

A 2.0- kg block is attached to a spring that has a force constant of 25 n/m. The spring is stretched 0.40 m from its equilibrium position and released. Find the total energy of the system and the frequency of oscillation according to classical calculations. Assuming that the energy is quantized, find the quantum number n for the system oscillating with this amplitude.

Answer 1

Answer

given,

mass of block = 2 kg

force constant = 25 n/m

spring stretched = 0.4 m

energy of the system =$\dfrac{1}{2}kx^2$

                                    =$\dfrac{1}{2}\times 25 \times 0.4^2$

                                    = 2 N/m

frequency of the oscillator

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{25}{2}}$

f = 0.56 Hz

b) energy of quantum state

 E = n h f

$n= \dfrac{E}{hf}$

$n= \dfrac{2}{6.62\times 10^{-34}\times 0.56}$

$n = 0.53 \times 10^{34}$

c) Energy by one quantum state is

  E = n h f

n =1

$E = (6.62 \times 10^{-34})(0.56)$

$E = 3.73 \times 10^{-34}$