The problem describes the motion of a spring subject to damping or frictional forces, modeled by a function of the form:

$s(t) = 2e^{-1.7t} \sin(2\pi t)$

where $s(t)$ is the position of the point on the spring in centimeters, and $t$ is the time in seconds. We are tasked with finding the velocity after $t$ seconds.

Step 1: Differentiate to Find Velocity

The velocity $v(t)$ is the derivative of the position function $s(t)$ with respect to time $t$. To do this, we'll apply the product rule and chain rule.

The product rule states: $\frac{d}{dt}[f(t)g(t)] = f'(t)g(t) + f(t)g'(t)$

Here, $f(t) = 2e^{-1.7t}$ and $g(t) = \sin(2\pi t)$.

Step 2: Derivatives

• $f'(t) = 2 \cdot (-1.7) e^{-1.7t} = -3.4 e^{-1.7t}$
• $g'(t) = 2\pi \cos(2\pi t)$

Step 3: Applying the Product Rule

Using the product rule: $v(t) = \frac{d}{dt}[2e^{-1.7t} \sin(2\pi t)] = (-3.4 e^{-1.7t}) \sin(2\pi t) + (2e^{-1.7t})(2\pi \cos(2\pi t))$

Simplify this expression: $v(t) = -3.4 e^{-1.7t} \sin(2\pi t) + 4\pi e^{-1.7t} \cos(2\pi t)$

Thus, the velocity function is: $v(t) = e^{-1.7t} \left( -3.4 \sin(2\pi t) + 4\pi \cos(2\pi t) \right)$

This is the expression for the velocity at any time $t$.

Would you like further details or have any questions?

Related Questions:

1. What is the velocity at a specific time, such as $t = 1$ second?
2. How would the velocity change if the damping constant (1.7) is increased?
3. How does the oscillatory motion (given by the sine term) affect the overall behavior of the spring?
4. Can we compute the maximum velocity the spring reaches during its motion?
5. How does the velocity change as $t \to \infty$, and why?

Tip: For oscillating systems, the exponential decay function $e^{-at}$ typically models damping, reducing the amplitude of the motion over time.