Let's solve this problem step-by-step.

Part (a)

We need to find the expression for the potential energy $U$ of a mass on a spring.

The potential energy $U$ of a spring is given by: $U = \frac{1}{2} k x^2$ where $k$ is the spring constant and $x$ is the displacement from the equilibrium position.

Given: $x = A \cos(\omega t) + B \sin(\omega t)$ we substitute $x$ into the expression for $U$: $U = \frac{1}{2} k \left( A \cos(\omega t) + B \sin(\omega t) \right)^2$ Expanding this expression: $U = \frac{1}{2} k \left( A^2 \cos^2(\omega t) + 2AB \cos(\omega t) \sin(\omega t) + B^2 \sin^2(\omega t) \right)$ Using trigonometric identities:

• $\cos^2(\omega t) + \sin^2(\omega t) = 1$
• $2 \cos(\omega t) \sin(\omega t) = \sin(2 \omega t)$

The expression for $U$ becomes: $U = \frac{1}{2} k \left( A^2 \cos^2(\omega t) + B^2 \sin^2(\omega t) + AB \sin(2 \omega t) \right)$

Part (b)

We need to show that the mean value of $U$, denoted $\langle U \rangle$, is given by: $\langle U \rangle = \frac{k}{4} (A^2 + B^2)$

To find the mean value $\langle U \rangle$, we average $U$ over a full cycle. The terms involving $\cos^2(\omega t)$ and $\sin^2(\omega t)$ each average to $\frac{1}{2}$, while the $\sin(2 \omega t)$ term averages to zero over a full cycle.

Thus: $\langle U \rangle = \frac{1}{2} k \left( A^2 \cdot \frac{1}{2} + B^2 \cdot \frac{1}{2} \right) = \frac{k}{4} (A^2 + B^2)$

This confirms the given result.

Would you like further clarification on any part? Here are some related questions to expand on the topic:

1. What would the expression for the kinetic energy of this system look like?
2. How does the potential energy vary with time for this oscillating system?
3. Can we derive the total energy of the system and show that it remains constant?
4. How would damping affect the potential and kinetic energy in this oscillatory system?
5. What is the effect of changing the amplitude $A$ or $B$ on the mean potential energy?

Tip: In simple harmonic motion, energy oscillates between potential and kinetic forms while the total energy remains constant if there is no damping.