A spring is attached to the ceiling and pulled 12 cm down from equilibrium and released. The amplitude decreases by 8% each second. The spring oscillates 17 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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To find an equation for the distance, D, the end of the spring is below equilibrium in terms of seconds, t, we can start by finding the equation for the amplitude decay.
Given that the amplitude decreases by 8% each second, we can express it as:
A(t) = A(0) * (1 - 0.08)^t
Where A(t) is the amplitude at time t, A(0) is the initial amplitude, and t is the time in seconds.
Next, let's find the period, T, of the oscillation. The period is the time it takes for the spring to complete one full oscillation. Given that the spring oscillates 17 times each second, we can express the period as:
T = 1 / 17
Now, we can use the amplitude and period to find the equation for the distance, D, below equilibrium.
D(t) = A(t) * sin(2π * t / T)
This equation represents the displacement of the spring below equilibrium at time t. The sin function models the oscillation, and the amplitude A(t) accounts for the decay over time.
To find the equation for the distance D in terms of time, we can break down the problem into different components.
1. Period:
The number of oscillations per second is given as 17. The period, T, is the time it takes for one complete oscillation. Therefore, T = 1 / (number of oscillations per second) = 1 / 17 seconds.
2. Amplitude:
The amplitude decreases by 8% each second. This means that after 1 second, the amplitude will be 92% of its original value. In other words, if A0 is the initial amplitude, the amplitude A at time t is given by A = A0 * (0.92)^t.
3. Angular Frequency:
The angular frequency, ω, can be calculated using the formula ω = 2π / T, where T is the period.
4. Equation for Distance:
The equation for the distance D in terms of time t is given by D = A * cos(ωt), where A is the amplitude and ω is the angular frequency.
Combining these components, we can create the equation for the distance D:
D = (A0 * (0.92)^t) * cos((2π / T) * t)
Therefore, the equation for the distance D the end of the spring is below equilibrium in terms of seconds t is D = (A0 * (0.92)^t) * cos((2π / (1/17)) * t) or simplifying it further,
D = (A0 * (0.92)^t) * cos(34πt)
since the period is 1/17 sec, then we will have sin/cos (34πt). Since we are starting at the minimum, and cos(x) starts at the maximum,
D(t) = -12*0.92^(t) cos(34πt)
or, in the more usual form,
D(t) = -12 e^(-0.083t) cos(34πt)