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 given by the formula:
\[ E = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant and \( A \) is the amplitude.

2. **Determine the change in amplitude**:
We are given that the amplitude decreases by 4% during each cycle. This can be expressed as:
\[ A' = A - 0.04A = 0.96A \]
where \( A' \) is the new amplitude after one cycle.

3. **Calculate the new energy after the decrease in amplitude**:
The new mechanical energy \( E' \) after the decrease in amplitude is:
\[ E' = \frac{1}{2} k (A')^2 = \frac{1}{2} k (0.96A)^2 = \frac{1}{2} k (0.9216A^2) = 0.9216 \cdot \frac{1}{2} k A^2 = 0.9216 E \]

4. **Determine the energy lost**:
The energy lost during one cycle can be calculated as:
\[ \Delta E = E - E' = E - 0.9216E = (1 - 0.9216)E = 0.0784E \]

5. **Calculate the percentage of energy lost**:
To find the percentage of mechanical energy lost, we can express it as:
\[ \text{Percentage loss} = \left(\frac{\Delta E}{E}\right) \times 100 = \left(\frac{0.0784E}{E}\right) \times 100 = 7.84\% \]

### Final Answer:
The percentage of mechanical energy lost in each cycle is approximately **7.84%**.

---