Energy in simple harmonic motion (5 exercises)

Certainly! Let's go through each problem:

### Problem 6
For a mass-spring oscillator:

- **Given:**
- Amplitude \(A = 0.1 \, \text{m}\)
- Mass \(m = 0.2 \, \text{kg}\)
- Spring constant \(k = 50 \, \text{N/m}\)
- Displacement \(x = 0.05 \, \text{m}\)

**Total Energy \(E\)** is given by:

\[
E = \frac{1}{2} k A^2
\]

\[
E = \frac{1}{2} \times 50 \times (0.1)^2 = 0.25 \, \text{J}
\]

**Potential Energy \(U\)** at any displacement \(x\) is:

\[
U = \frac{1}{2} k x^2
\]

\[
U = \frac{1}{2} \times 50 \times (0.05)^2 = 0.0625 \, \text{J}
\]

**Kinetic Energy \(K\)** is:

\[
K = E - U
\]

\[
K = 0.25 - 0.0625 = 0.1875 \, \text{J}
\]

### Problem 7
**Show that total energy is proportional to the square of amplitude:**

The total energy \(E\) in simple harmonic motion for a mass-spring system is:

\[
E = \frac{1}{2} k A^2
\]

This equation demonstrates that \(E\) is proportional to \(A^2\) since \(k\) is constant.

### Problem 8
**Given:**
- Kinetic Energy \(K = 0.045 \, \text{J}\)
- Displacement \(x = 0.06 \, \text{m}\)
- Mass \(m = 0.3 \, \text{kg}\)

We first find \(k\) using the total energy.

**Potential Energy \(U\):**
\[
U = \frac{1}{2} k x^2
\]

Total Energy \(E\) is the sum of \(K\) and \(U\):

\[
E = K + U
\]

Total \(U = E - K\).

Using mass and speed at equilibrium:
\[
K = \frac{1}{2} m v^2
\]
At \(x = 0.06\), find:
\[
v = \omega \sqrt{A^2 - x^2}
\]

### Problem 9
**Amplitude tripled effect on energy:**

If the amplitude \(A\) is tripled, total energy \(E\) becomes:

\[
E \propto A^2
\]

If \(A\) becomes \(3A\), then \(E\) becomes:

\[
E \propto (3A)^2 = 9A^2
\]

Thus, the total energy increases by a factor of 9.

These solutions will help you understand energy transformation and conservation in harmonic systems.