# Video: Quantum Energy Number of a Macroscopic Object

A particle with mass 0.0300 kg oscillates back-and-forth on a spring with frequency 4.00 Hz. At the spring’s equilibrium position, the particle has a speed of 0.600 m/s. If the particle is in a state of definite energy, find its energy quantum number.

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### Video Transcript

A particle with mass 0.0300 kilograms oscillates back-and-forth on a spring with frequency 4.00 hertz. At the spring’s equilibrium position, the particle has a speed of 0.600 meters per second. If the particle is in a state of definite energy, find its energy quantum number.

Let’s start by recording some of this information given. We’re told the mass of the particle is 0.0300 kilograms, which we’ll call 𝑚. The frequency of oscillation of the particles 4.00 hertz or 𝑓. When the spring the particle is on is at equilibrium, neither stretched nor compressed, this particle has a speed of 0.600 meters per second, what we’ll call 𝑣.

We’ll start out by drawing a sketch of this situation. Since we’re seeking to solve for the energy quantum number of the particle in a state of definite energy, which we’ll call capital 𝑁, in our sketch, our mass is on a string moving back and forth with a frequency 𝑓. When the mass is at its equilibrium position, it moves with a speed 𝑣 of 0.600 meters per second. Along with 𝑓 and 𝑣, we’re given the object’s mass 𝑚. Since we’re told this object is in a definite state of energy and has a quantum number and is oscillating back and forth, we think of the quantum harmonic oscillator.

The energy of an object in the 𝑛th state of a quantum harmonic oscillator is 𝑛 plus a half times Planck’s constant times the frequency of oscillation. In our case, 𝐸 sub capital 𝑁 equals capital 𝑁, the quantum number we wanna solve for, plus a half times Planck’s constant times frequency. Now we’re going to link the energy of this object with its speed and we’ll do that by using the work-energy theorem. Recall that this theorem says that the work done on an object in units of joules equals its change in kinetic energy. In our case, this work 𝑊 is energy provided by the spring which we can call 𝐸 sub s. And Δ𝐾𝐸, we can recall is equal to the change in an object’s speed squared times its mass times a half.

So the energy of our oscillator is equal to the energy provided by the spring which, by the work-energy theorem, is itself equal to one-half the mass times the speed of the mass squared. All this means that capital 𝑁 plus a half times Planck’s constant times ℎ equals one-half 𝑚𝑣 squared.

When we rearrange this expression to solve for capital 𝑁, we find it’s equal to one over two ℎ𝑓 times the quantity 𝑚𝑣 squared minus ℎ𝑓. In this problem, we’ll treat ℎ as exactly 6.626 times 10 to the negative 34th joule seconds. We’re now ready to plug in for ℎ, 𝑓, 𝑚 and 𝑣 to solve for capital 𝑁.

When we do — plugging in for ℎ, for 𝑚, for 𝑣, and for 𝑓, all in SI units — and enter these values on our calculator, we find that, to three significant figures, 𝑁 is 2.03 times 10 to the 30th. That’s the quantum number of an oscillator with this energy.