Physics, 05.10.2019 02:30 daphnewibranowsky

Amass of 3 kg is attached to a spring hanging under gravity with spring con- stant 12. if friction is negligible and the mass is pulled down 1 metre from the equilibrium position and released without kick, find the position of the mass at any later time t. (b) assume damping, with damping constant d, is introduced into the system given in part (a) so that damping is critical. find d.

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Answer from: kiki4832

a) $y(t) = sin(2t)+\frac{gt^{2} }{2}$

b) "d" value is 12.

Explanation:

Let us set our equation for the general movement:

$m\frac{d^{2}y }{dt^{2} } +d\frac{dy }{dt } +ky=mg$

Where y(t) stands for the vertical displacement. Our first term stands for the acceleration/innertia, the second one is the damping term, the third it the spring force (proportional to the vertical displacement due to the spring formula) and the other side of the equation is the constant force (weight). All in all, we got nothing than the Second Newton's Law.

Let's cancel the damping term for question a) ("friction is negligible"). The equation remains as follows:

$m\frac{d^{2}y }{dt^{2} } +ky=mg$

This is an ordinary differential equation or ODE, with it's homogeneous and particular solutions.

The homogeneous solutions is $y_{h} (t) = A sin(wt)$ , where ω is the natural frequency of the system:

$w=\sqrt{\frac{k}{m} } =\sqrt{\frac{12N/m\\}{3kg} } =\sqrt{4} =2 rad/s$

The particular solutions results from integrating the equation by ignoring the spring term, since $y (t) = y_{h} (t) + y_{p} (t)$ , and $y_{h} (t)$ would be cancelled by the spring term. Thus:

$y_{p} (t) = g\frac{t^{2} }{2}$

So, the whole solution would be:

$y (t) = y_{h} (t) + y_{p} (t) = A sin(2t)+g\frac{t^{2} }{2}$ .

A is a constant to be set from the initial conditions:

y (t=0) = -1 = A (assuming the spring is initially stretched downwards.

Then for any time "t", $y (t) = -sin(2t)+g\frac{t^{2} }{2}$ .

Remember g = 9,81 m/s2, it is a constant, known value.

b) Returning to our former, complete equation, and applying the known values:

$3\frac{d^{2}y }{dt^{2} } +d\frac{dy }{dt } +12y=3g$ .

Here, d=2jmw=12j , where "j" is the damping parameter. A critical damping takes place when "j" value is equal to 1, so "d" value (the whole term's multiplier) must be equal to 12.

Another way to solve this part is to obtain the caracteristic polinom for the damping oscilator:

$x^{2} +2jwx+w^{2} =0$

This polinom is useful when you want to find the critical damping value, that is, when the above equation only has one (double) solution for "x", regardless the value of frequency "x". This hapens for j=1.

Always bare in mind that the complete damping term can be written as 2jmwy(t) in the equation.

Answer from: Quest

high-sugar diet and high cholesterol

Answer from: Quest

force required to detach it is given by

$f = 4 * 10^{-8} n$

charges are taken to be fundamental charge

$q_1 = q_2 = 1.6 * 10^{-19}c$

now we have

$f = \frac{kq_1q_2}{r^2}$

$4 * 10^{-8} = \frac{9*10^9 * 1.6*10^{-19}* 1.6* 10^{-19}}{r^2}$

$r = 7.6 * 10^{-11} m$

so above is the minimum distance at which it will detach

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