Finding an equation for a spring which is pulled down from equilibrium (solution-verification)

[Stallmp]

Let's suppose that we have a spring which is attached to the ceiling and pulled 12 cm down from equilibrium and released. After 2 seconds, the amplitude has decreased to 7 cm. The spring oscillates 11 times each second. Assume that the amplitude is decreasing exponentially. Now we have to find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

Since the spring at t=0 is pulled 12 cm down from equilibrium, we know that it starts at a minimum value. So it is a cosine function with a negative amplitude. The frequency is 11 Hz since the spring oscillates 11 times each second, so b (the value we enter in the formula) is [MATH]2\pi*11 = 22\pi[/MATH] (because the period is [MATH]1/11[/MATH]). The amplitude represents an exponential function, which after 2 seconds decreases to 7 cm. This means that [MATH]7 = 12 * g^2 => g = \sqrt\frac{7}{12}[/MATH]. So the function I end up with is [MATH]D(t) = -12 * (\sqrt\frac{7}{12})^t*cos(22\pi t)[/MATH]. For some reason my answer has been marked as wrong. Can someone please point out my mistake?

 

[Deleted member 4993]

Let's suppose that we have a spring which is attached to the ceiling and pulled 12 cm down from equilibrium and released. After 2 seconds, the amplitude has decreased to 7 cm. The spring oscillates 11 times each second. Assume that the amplitude is decreasing exponentially. Now we have to find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

Since the spring at t=0 is pulled 12 cm down from equilibrium, we know that it starts at a minimum value. So it is a cosine function with a negative amplitude. The frequency is 11 Hz since the spring oscillates 11 times each second, so b (the value we enter in the formula) is [MATH]2\pi*11 = 22\pi[/MATH] (because the period is [MATH]1/11[/MATH]). The amplitude represents an exponential function, which after 2 seconds decreases to 7 cm. This means that [MATH]7 = 12 * g^2 => g = \sqrt\frac{7}{12}[/MATH]. So the function I end up with is [MATH]D(t) = -12 * (\sqrt\frac{7}{12})^t*cos(22\pi t)[/MATH]. For some reason my answer has been marked as wrong. Can someone please point out my mistake?

Are you trying to model a "damped simple harmonic oscillator"?

Please post the "exact" problem - and its context (homework or project or work-assignment, etc.).

The general equation is:

x(t)=Ae^−gte^iat

For a quick reference, please read:

Damped Harmonic Oscillators | Brilliant Math & Science Wiki

Damped harmonic oscillators are vibrating systems for which the amplitude of vibration decreases over time. Since nearly all physical systems involve considerations such as air resistance, friction, and intermolecular forces where energy in the system is lost to heat or sound, accounting for...

brilliant.org

 

[Stallmp]

"
Let's suppose that we have a spring which is attached to the ceiling and pulled 12 cm down from equilibrium and released. After 2 seconds, the amplitude has decreased to 7 cm. The spring oscillates 11 times each second. Assume that the amplitude is decreasing exponentially. Now we have to find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

" - This is like 100% the exact problem. It's a mathematical problem which involves trigonometry. You don't need to know any physics for this problem. This section falls under trigonometric functions, such as cosine and sine and simply how to construct functions. So this is considered to be a pretty basic high school trigonometry problem.

So I am simply wondering if you see a mistake in my answer and if yes, what the correct answer should be.

 

[Dr.Peterson]

Since the spring at t=0 is pulled 12 cm down from equilibrium, we know that it starts at a minimum value. So it is a cosine function with a negative amplitude. The frequency is 11 Hz since the spring oscillates 11 times each second, so b (the value we enter in the formula) is [MATH]2\pi*11 = 22\pi[/MATH] (because the period is [MATH]1/11[/MATH]). The amplitude represents an exponential function, which after 2 seconds decreases to 7 cm. This means that [MATH]7 = 12 * g^2 => g = \sqrt\frac{7}{12}[/MATH]. So the function I end up with is [MATH]D(t) = -12 * (\sqrt\frac{7}{12})^t*cos(22\pi t)[/MATH]. For some reason my answer has been marked as wrong. Can someone please point out my mistake?

It would be very helpful to know more about what you have been taught about this sort of motion, because you may have been given an oversimplified version of the equation.

One potential mistake you made is in putting the given (2, 7) into the pure exponential form rather than the actual equation you are using for the motion. That happens not to cause trouble because 2 seconds is a multiple of the period, so you do get [MATH]D(2) = -12 \left(\sqrt\frac{7}{12}\right)^2\cos(22\pi\cdot2) =-7\cos(44\pi) = -7[/MATH]. If "amplitude" is taken to mean the number multiplying the cosine, rather than the actual distance, or if you paid attention to where that point is in the cycle, then you did nothing wrong there.

Another possible mistake, depending on what you have been taught, is that in your function, t=0 is not, in fact, a minimum! That's because the cosine is being multiplied by the exponential, so that the slope at that point is not exactly zero. But this is a very small error, and you may have been taught to ignore it, if you are not doing any calculus.

Possibly what was marked wrong is merely the sign. The equation asks for "the distance, D the end of the spring is below equilibrium". This is a positive number when it is below, not negative as you have it. So just remove the negative in your answer.

Was this marked by a human (who should be able to point out how close you are, and answer questions about the error), or by a computer?