A spring is attached to the ceiling and pulled 10 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 9 times each second. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

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A spring is attached to the ceiling and pulled 10 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 9 times each second. Find an equation for the distance, \( D \), the end of the spring is below equilibrium in terms of seconds, \( t \).

\[ D(t) = \]

Explanation:

This problem involves the motion of a damped harmonic oscillator. The spring's initial displacement, amplitude decay, and oscillation frequency create a scenario where you can model the motion mathematically.

1. **Initial Conditions**: 
   - Initial displacement: 10 cm
   - Amplitude decay per second: 11%
   - Frequency of oscillation: 9 cycles per second

2. **Equation Components**:
   - **Amplitude Decay**: The exponential decay formula is used for reducing amplitude. If the amplitude decreases by 11% each second, the decay factor per second is \( 1 - 0.11 = 0.89 \).

   - **Oscillation**: The oscillation can be described using a sine or cosine function. The angular frequency \( \omega \) is calculated from the frequency as \( \omega = 2\pi \times \text{frequency} \).

3. **General Form of the Equation**:
   - The equation is in the form \( D(t) = A_0 \times (0.89)^t \times \cos(\omega t) \) or using sine, depending on the initial phase.

4. **Calculating Components**:
   - Initial Amplitude (\( A_0 \)): 10 cm
   - Amplitude reduction factor per second: 0.89
   - Frequency: 9 Hz, so \( \omega = 2\pi \times 9 \)

5. **Final Equation**:
   - \( D(t) = 10 \times (0.89)^t \times \cos(18\pi t) \)

The equation \( D(t) \) models the distance of the spring's end below equilibrium at any time \( t \), accounting for both the oscillatory nature and the damping effect from amplitude reduction.
Transcribed Image Text:A spring is attached to the ceiling and pulled 10 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 9 times each second. Find an equation for the distance, \( D \), the end of the spring is below equilibrium in terms of seconds, \( t \). \[ D(t) = \] Explanation: This problem involves the motion of a damped harmonic oscillator. The spring's initial displacement, amplitude decay, and oscillation frequency create a scenario where you can model the motion mathematically. 1. **Initial Conditions**: - Initial displacement: 10 cm - Amplitude decay per second: 11% - Frequency of oscillation: 9 cycles per second 2. **Equation Components**: - **Amplitude Decay**: The exponential decay formula is used for reducing amplitude. If the amplitude decreases by 11% each second, the decay factor per second is \( 1 - 0.11 = 0.89 \). - **Oscillation**: The oscillation can be described using a sine or cosine function. The angular frequency \( \omega \) is calculated from the frequency as \( \omega = 2\pi \times \text{frequency} \). 3. **General Form of the Equation**: - The equation is in the form \( D(t) = A_0 \times (0.89)^t \times \cos(\omega t) \) or using sine, depending on the initial phase. 4. **Calculating Components**: - Initial Amplitude (\( A_0 \)): 10 cm - Amplitude reduction factor per second: 0.89 - Frequency: 9 Hz, so \( \omega = 2\pi \times 9 \) 5. **Final Equation**: - \( D(t) = 10 \times (0.89)^t \times \cos(18\pi t) \) The equation \( D(t) \) models the distance of the spring's end below equilibrium at any time \( t \), accounting for both the oscillatory nature and the damping effect from amplitude reduction.
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