A spring is attached to the ceiling and pulled 6 cm down from equilibrium and released. After 4 seconds the amplitude has decreased to 3 cm. The spring oscillates 15 times each second. Assume that the amplitude is decreasing exponentially. Find an equation for the distance, D the end of the spring is below equilibrium in terms of second
To find the equation for the distance D, the end of the spring is below equilibrium in terms of time, we can start by finding the damping constant, b, using the given information.
The amplitude of an exponentially decreasing oscillation can be modeled by the equation:
A(t) = A0 * e^(-bt)
Where A(t) is the amplitude at time t, A0 is the initial amplitude, b is the damping constant, and e is Euler's number, approximately 2.71828.
In this case, we are given that the initial amplitude A0 is 6 cm and the amplitude after 4 seconds A(4) is 3 cm.
We can substitute these values into the equation and solve for b:
3 = 6 * e^(-4b)
Divide both sides by 6:
0.5 = e^(-4b)
Take the natural logarithm (ln) of both sides to isolate the exponent:
ln(0.5) = -4b
Now solve for b by dividing both sides by -4:
b = ln(0.5) / -4
Using a calculator, we can find that b is approximately 0.1733.
Now that we have the damping constant, we can find the equation for the distance D below equilibrium in terms of time.
Let's assume the equilibrium position is at y = 0. When the oscillation is at maximum displacement, the distance D below equilibrium is equal to the amplitude A(t):
D = A(t)
Substituting the equation for the amplitude we derived earlier:
D = A0 * e^(-bt)
Plugging in the values of A0 = 6 cm and b = 0.1733:
D = 6 * e^(-0.1733t)
Therefore, the equation for the distance D the end of the spring is below equilibrium in terms of time t is D = 6 * e^(-0.1733t).