Question 40.2: The Quantized Oscillator A 2.00-kg block is attached to a ma......

(A) Conceptualize We understand the details of the block’s motion from our study of simple harmonic motion in Chapter 15. Review that material if you need to.

Categorize The phrase “according to classical calculations” tells us to categorize this part of the problem as a classical analysis of the oscillator. We model the block as a particle in simple harmonic motion.

Analyze Based on the way the block is set into motion, its amplitude is 0.400 \mathrm{~m}.

Evaluate the total energy of the block-spring system using Equation 15.21:

E=\frac{1}{2} k A^{2}=\frac{1}{2}(25.0 \mathrm{~N} / \mathrm{m})(0.400 \mathrm{~m})^{2}=2.00 \mathrm{~J}

Evaluate the frequency of oscillation from Equation 15.14:

f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{25.0 \mathrm{~N} / \mathrm{m}}{2.00 \mathrm{~kg}}}=0.563 \mathrm{~Hz}

(B) Categorize This part of the problem is categorized as a quantum analysis of the oscillator. We model the block-spring system as a Planck oscillator.

Analyze Solve Equation 40.4

E_{n}=n h f        (40.4)

for the quantum number n :

n=\frac{E_{n}}{h f}

Substitute numerical values:

n=\frac{2.00 \mathrm{~J}}{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)(0.563 \mathrm{~Hz})}=5.36 \times 10^{33}

Finalize Notice that 5.36 \times 10^{33} is a very large quantum number, which is typical for macroscopic systems. Changes between quantum states for the oscillator are explored next.