A spring is attached to the ceiling and pulled 17 cm down from equilibrium and released. After 3 seconds the amplitude has decreased to 16 cm. The spring oscillates 13 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 seconds, t.
Why did the spring go to a therapist? Because it had some serious oscillation issues!
Now, let's bounce into your question. We know that the amplitude of the spring's oscillation decreases exponentially over time. So, let's call the initial amplitude A₀ and the final amplitude Aₜ (after time t). In this case, A₀ = 17 cm and Aₜ = 16 cm.
The general equation for exponential decay can be written as:
Aₜ = A₀ * e^(-kt),
where k is the decay constant.
Since we're given that the spring oscillates 13 times per second, which means it completes 13 cycles in one second, the frequency of oscillation (f) can be calculated as:
f = 13 cycles/second.
Now, the period (T) of one complete oscillation is the reciprocal of the frequency:
T = 1/f = 1/13 seconds.
We can relate the decay constant (k) to the period (T) using the equation:
k = 2π / T.
Plugging in the known values, we get:
k = 2π / (1/13).
Now, let's substitute this value of k into our exponential decay equation:
Aₜ = A₀ * e^(- (2π / T) * t).
Simplifying this equation, we have:
Aₜ = A₀ * e^(- (26π) * t).
Now, we have an equation relating the amplitude (Aₜ) to time (t) as the spring oscillates. But you asked for the distance, D, the end of the spring is below equilibrium.
Since the amplitude of the spring's oscillation is equal to half of the distance D, we can write:
D = 2 * Aₜ.
Substituting the equation we derived for Aₜ into this equation, we get:
D = 2 * (A₀ * e^(- (26π) * t)).
So, the equation for the distance, D, the end of the spring is below equilibrium in terms of time, t, is:
D = 2 * (17 cm) * e^(- (26π) * t).
Remember, this equation assumes that the amplitude of the spring's oscillation decreases exponentially over time.
To find an equation for the distance, D, the end of the spring is below equilibrium in terms of seconds, t, we can use the formula for simple harmonic motion:
D = A * e^(-kt) * cos(ωt + φ)
Where:
- D is the distance the end of the spring is below equilibrium,
- A is the initial amplitude of the spring,
- e is the mathematical constant approximately equal to 2.71828,
- k is the decay constant,
- t is the time in seconds,
- ω is the angular frequency (2π times the frequency),
- φ is the phase constant.
Given:
- The spring is pulled 17 cm down from equilibrium, so A = 17 cm.
- After 3 seconds, the amplitude has decreased to 16 cm.
- The spring oscillates 13 times each second, so the frequency is 13 Hz.
To find the decay constant, k, we can use the information about the amplitude decreasing from 17 cm to 16 cm over 3 seconds:
A * e^(-k * 3) = 16
Dividing both sides by 17 cm:
e^(-k * 3) = 16/17
Taking the natural logarithm (ln) of both sides:
-ln(17/16) = -k * 3
Simplifying:
k = ln(17/16) / 3
Since ω = 2πf and the frequency is 13 Hz, ω = 2π * 13 = 26π rad/s.
Assuming the phase constant φ is 0, the equation becomes:
D = 17 * e^(-kt) * cos(26πt)
So, the equation for the distance, D, the end of the spring is below equilibrium in terms of seconds, t, is:
D = 17 * e^(ln(17/16)/3 * t) * cos(26πt)
To find an equation for the distance, D, at the end of the spring below equilibrium in terms of seconds, t, we need to use the information given.
First, let's define some variables:
- A: Initial amplitude (17 cm)
- t: Time (in seconds)
- A_0: Amplitude at time t = 0 (17 cm)
- A_t: Amplitude at time t (in cm)
We are given that the amplitude decreases exponentially. This means we can use the formula for exponential decay:
A_t = A_0 * e^(-kt)
Where k is a constant related to the rate at which the amplitude decreases.
Now, we know that after 3 seconds, the amplitude has decreased to 16 cm. Substituting these values into the equation, we have:
16 = 17 * e^(-k*3)
To find the value of k, let's solve this equation for k. Divide both sides by 17:
16/17 = e^(-k*3)
Now, take the natural logarithm (ln) of both sides to remove the exponential:
ln(16/17) = -k*3
Divide both sides by -3:
k = -ln(16/17) / 3
Now that we have the value of k, we can proceed to find the equation for the distance, D, at time t. The distance, D, is equal to the amplitude, A_t, multiplied by the sine function of the frequency, f, multiplied by t:
D = A_t * sin(2πf*t)
Given that the spring oscillates 13 times per second, the frequency, f, is 13 times 2π:
f = 13 * 2π
Substituting the value of A_t from the exponential decay equation, we have:
D = (A_0 * e^(-kt)) * sin(2πf*t)
Substituting the previously calculated value of k, we get:
D = (17 * e^(-(-ln(16/17) / 3)*t)) * sin(2πf*t)
Simplifying further, we have:
D = 17 * e^(ln(16/17) * t/3) * sin(26πt)
Therefore, the equation for the distance, D, at the end of the spring below equilibrium in terms of seconds, t, is:
D = 17 * (16/17)^(t/3) * sin(26πt)
since the spring starts at (0,-17), we know that
D(t) = -17 e^-at cos(kt)
since the period is 1/13, k = 26π and we have
D(t) = -17 e^-at cos(26πt)
Now, we know that the amplitude has decreased to (16/17)^(t/3)
That means that a = -1/3 ln(16/17) = .02, and
D(t) = -17 e^-0.02t cos(26πt)
You can see a plot at
http://www.wolframalpha.com/input/?i=plot+y+%3D+-17+e^%28-0.02t%29+cos%2826%CF%80t%29%2C+y%3D16+for+t%3D0+to+3