The total energy of the oscillations is 40 J when the amplitude of the oscillations is
0.50 m. Sketch a graph showing how the total energy of the child varies with the
amplitude of the oscillations for amplitudes between 0 and 1.00 m. Include a suitable
scale on the total energy axis.

To solve this problem, we start by analyzing the relationship between total energy and amplitude for oscillations. The total mechanical energy \( E \) in simple harmonic motion is proportional to the square of the amplitude \( A \). This can be described by the equation:

\[
E \propto A^2
\]

Or, more specifically,

\[
E = kA^2
\]

where \( k \) is a constant of proportionality that depends on the mass and spring constant or other system parameters.

### Step 1: Use the given information
We know that:
- The total energy \( E = 40 \, \text{J} \) when the amplitude \( A = 0.50 \, \text{m} \).

Substitute these values into the equation to find the constant \( k \):

\[
40 = k(0.50)^2
\]
\[
40 = k(0.25)
\]
\[
k = \frac{40}{0.25} = 160
\]

Thus, the relationship between energy and amplitude is:

\[
E = 160A^2
\]

### Step 2: Plot the equation
The equation \( E = 160A^2 \) is a quadratic equation, where the total energy increases with the square of the amplitude.

#### Energy values at various amplitudes:
- For \( A = 0 \, \text{m} \), \( E = 160(0)^2 = 0 \, \text{J} \)
- For \( A = 0.25 \, \text{m} \), \( E = 160(0.25)^2 = 160(0.0625) = 10 \, \text{J} \)
- For \( A = 0.50 \, \text{m} \), \( E = 40 \, \text{J} \) (as given)
- For \( A = 0.75 \, \text{m} \), \( E = 160(0.75)^2 = 160(0.5625) = 90 \, \text{J} \)
- For \( A = 1.00 \, \text{m} \), \( E = 160(1.00)^2 = 160 \, \text{J} \)

### Step 3: Graph the relationship
Now, you can sketch the graph of \( E = 160A^2 \). The amplitude \( A \) is on the x-axis (from 0 to 1.00 m), and the total energy \( E \) is on the y-axis. The y-axis should be scaled to go from 0 to at least 160 J. The graph will be a parabola opening upwards, starting at \( E = 0 \) when \( A = 0 \), passing through \( E = 40 \, \text{J} \) at \( A = 0.50 \, \text{m} \), and reaching \( E = 160 \, \text{J} \) at \( A = 1.00 \, \text{m} \).

Would you like me to generate this graph for you?

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### 5 Related Questions:
1. How does the energy change if the amplitude of oscillation is doubled?
2. What happens to the total energy if the mass of the child is doubled, keeping amplitude constant?
3. How is the constant \( k \) affected by changes in the spring constant or length of a pendulum?
4. How would the energy vs. amplitude graph differ for a damped oscillator?
5. What is the relationship between frequency and amplitude in simple harmonic motion?

**Tip:** The total energy in oscillatory systems is always proportional to the square of the amplitude for simple harmonic motion.

The total energy of the oscillations is 40 J when the amplitude of the oscillations is 0.50 m. Sketch a graph showing how the total energy of the child varies with the amplitude of the oscillations for amplitudes between 0 and 1.00 m. Include a suitable scale on the total energy axis.

Explore how total energy in simple harmonic motion varies with amplitude. Using the relationship E = kA^2, learn to calculate energy at different amplitudes and plot a quadratic graph. Suitable for high school students studying oscillatory systems and energy in physics.

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