A spring attached to the ceiling is pulled 8 cm down from cquilibrium and released. The amplitude decreases by 6% cach second. The spring oscillates 25 times each second. Find a function that models the distance, D, the end of the spring is from equilibrium in terms of seconds, t, since the spring was released. Assume that the spring is pulled from equilibrium in the positive direction.
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![**Modeling Oscillations of a Spring**
**Problem Description:**
A spring attached to the ceiling is pulled 8 cm down from equilibrium and released. The amplitude decreases by 6% each second. The spring oscillates 25 times each second. Find a function that models the distance, \( D \), the end of the spring is from equilibrium in terms of seconds, \( t \), since the spring was released. Assume that the spring is pulled from equilibrium in the positive direction.
**Mathematical Formulation:**
To find the function \( D(t) \), we need to account for both the oscillatory behavior and the decay in amplitude over time.
1. **Oscillatory Component:**
- The spring oscillates \( 25 \) times each second. This means the frequency \( f \) is \( 25 \) Hz.
- The angular frequency \( \omega \) can be calculated as \( \omega = 2\pi f = 2\pi \times 25 = 50\pi \) (radians per second).
- The initial displacement (amplitude) is \( 8 \) cm.
2. **Amplitude Decay:**
- The amplitude decreases by \( 6\% \) each second. This can be modeled by an exponential decay function.
- If the initial amplitude is \( A_0 = 8 \) cm, then the amplitude as a function of time \( t \) is \( A(t) = A_0 \times (1 - 0.06)^t = 8 \times (0.94)^t \).
Combining both components, the distance \( D \) from equilibrium at any time \( t \) is given by:
\[ D(t) = A(t) \cos(\omega t) = 8 \times (0.94)^t \cos(50\pi t) \]
**Function:**
\[ D(t) = 8 \cdot (0.94)^t \cdot \cos(50\pi t) \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F416064f9-c57b-46ee-9b50-14b257515654%2Fca8fc34e-39fc-47de-85b2-7bddc5c099fc%2Fg2jqr3g.png&w=3840&q=75)
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