Question

Simple harmonic oscillation occurs when the potential energy function is equal to \(\frac{1}{2} k x^{2},\) where \(k\) is a constant. What happens to the ground- state energy level if \(k\) is increased? a) It increases. b) It remain the same. c) It decreases.

Short answer

Answer: a) It increases.

Step-by-step solution

Identify the ground state energy level in simple harmonic oscillation

In a simple harmonic oscillator, the ground state energy level (lowest energy level) can be found using the formula \(E_0 = \frac{1}{2}\hbar\omega\), where \(\hbar\) is the reduced Planck's constant and \(\omega\) is the angular frequency. Next, let's find the angular frequency.

Calculate the angular frequency

The angular frequency can be determined using the relation \(\omega = \sqrt{\frac{k}{m}}\), where \(k\) is the spring constant and \(m\) is the mass of the oscillator. Now, we will substitute this expression for the angular frequency in the ground state energy formula.

Substitute the expression of angular frequency in the ground state energy formula

We substitute the expression for the angular frequency in the ground state energy formula: \(E_0 = \frac{1}{2}\hbar\omega = \frac{1}{2}\hbar\sqrt{\frac{k}{m}}\) Now, we'll analyze how the ground state energy depends on the value of \(k\).

Analyze the dependence of ground state energy on the spring constant

From the ground state energy expression, \(E_0 = \frac{1}{2}\hbar\sqrt{\frac{k}{m}}\), we can see that the ground state energy level is directly proportional to the square root of the spring constant \(k\). This means that if \(k\) increases, the energy level will also increase. Therefore, the correct answer is: a) It increases.