The amplitude of a lightly damped oscillator decreases by 4.0% during each cycle. What percentage of mechanical energy of the oscillator is lost in each cycle?

Explanation

To solve the problem of determining the percentage of mechanical energy lost in each cycle of a lightly damped oscillator, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between Amplitude and Energy**:
The mechanical energy (E) of a simple harmonic oscillator is proportional to the square of its amplitude (A). This can be expressed as:
\[ E \propto A^2 \]

2. **Identify the Percentage Decrease in Amplitude**:
The problem states that the amplitude decreases by 4% during each cycle. This means that if the initial amplitude is \( A \), the new amplitude after one cycle is:
\[ A' = A - 0.04A = 0.96A \]

3. **Calculate the Initial and Final Energies**:
The initial mechanical energy \( E_i \) when the amplitude is \( A \) is:
\[ E_i = kA^2 \]
where \( k \) is a proportionality constant. The final mechanical energy \( E_f \) when the amplitude is \( 0.96A \) is:
\[ E_f = k(A')^2 = k(0.96A)^2 = k(0.9216A^2) \]

4. **Determine the Change in Energy**:
The change in mechanical energy \( \Delta E \) can be calculated as:
\[ \Delta E = E_f - E_i = k(0.9216A^2) - k(A^2) = k(0.9216A^2 - A^2) = k(-0.0784A^2) \]

5. **Calculate the Percentage Change in Energy**:
The percentage change in energy is given by:
\[ \text{Percentage Change} = \frac{\Delta E}{E_i} \times 100 = \frac{-0.0784A^2}{A^2} \times 100 = -7.84\% \]
Since we are interested in the loss of energy, we take the absolute value:
\[ \text{Percentage Loss} = 7.84\% \]

6. **Final Result**:
Therefore, the percentage of mechanical energy lost in each cycle is approximately:
\[ \text{Percentage of Energy Lost} \approx 7.84\% \]