A 15-kg mass, attached to a massless spring whose force constant is 2500 N/m, has an amplitude of 4 cm. Assuming the energy is quantized, find the quantum number of the system, n, if E_n = nhf. A) 1.5 × 10³³ B) 3.0 × 10³³ C) 4.5 × 10³³ D) 5.4 × 10³³ E) 1.0 × 10³³

Question

Quizplus · Accepted Answer

First, we need to find the frequency of the system using the equation:

f = (1/2π) √(k/m)

f = (1/2π) √(2500/15) = 8.54 Hz

Next, we can find the energy using the equation:

E = (n + 1/2)hf

where h is Planck's constant, and n is the quantum number.

We need to convert the amplitude to meters:

4 cm = 0.04 m

The maximum potential energy of the system is when the mass is at the maximum displacement from equilibrium, which is given by:

Umax = (1/2)kA^2

Umax = (1/2)(2500)(0.04)^2 = 2 J

Therefore, the energy levels of the system are:

E1 = 0 J
E2 = (1/2)hf = 2 J
E3 = (3/2)hf = 4 J
E4 = (5/2)hf = 6 J
E5 = (7/2)hf = 8 J

We can see that the maximum potential energy corresponds to the second energy level, which means:

2 J = (n + 1/2)hf

Solving for n:

n = (2 J)/(hf) - 1/2 = 1.5 × 10^3

Therefore, the quantum number of the system is 1.5 × 10^3. The closest answer choice is A.

A 15-Kg Mass, Attached to a Massless Spring Whose Force