when we assume a spring is perfectly elastic, that spring obeys Hooke’s Law , wh

Question

when we assume a spring is perfectly elastic, that spring obeys Hooke’s Law, whch states that the force on a spring is proportional to the springs displacement x from equilibrium.

F = kx

Here, we have that

•F is a force given in Newtons

•k is the spring constant which is given in Newtons per meter (N/m)

•x is the distance (technically displacement) from the spring’s equilibrium.

The spring constant can be thought of as a measure of how ”stiff” a spring is; that is, how resistant a spring is to pushing and pulling from its equilibrium position. The bigger the value of k, the stiffer the spring.

For this type of problem, we will be dealing with a model of the form

x ( t ) = x cos r k m t + v r m k

Here, we have that

•u is the initial displacement from equilitbrium, measured in meters

•v is the initial velocity of the spring, measured in meters per second

–a positive initial velocity means we give the weight at the end of the spring an upward push, so the spring is compressed.

–a negative initial velocity means we give the weight at the end of the spring an downward push, so the spring is stretched.

•k is the spring constant in Newtons per meter

•m is the mass of the weight attached to the end of the spring in kilograms (kg)

As with the previous section, we will construct a number of simple models and then answer some questions about them. For these problems, we again assume our springs are perfectly elastic and we do not have to worry about friction. Consider the following scenarios.

1.Suppose that it takes a force of F = 20 newtons to compress a spring by 2 meters. Using a little algebra and Hooke’s Law (and neglecting the negative sign), what is our spring constant k in this case, and what are the units?

2.Using the spring constant from problem 1, suppose that a weight of 2 kg on the end of thespring starts out 0.5 meters above equilibrium with no initial push.

•Write out and sketch the model out on the interval [0,2].

•Based on your model, what is the amplitude? How would you interpret this physically?

•Based on your model, what is the period? How would you interpret this physically?

•Based on the graph of your model, what is the approximate phase shift?

•Using your previous answers, how could you rewrite your model (approximately) to include the phase shift?

3.Now suppose that you start off the previous spring with an initialial velocity of 2 meters persecond.

•Write out and sketch the model out on the interval [0,2].

•Based on the graph of your model, what is the approximate amplitude? How would you interpret this physically?

•Based on the graph of your model, what is the approximate period? How would you interpret this physically?

•Based on the graph of your model, what is the approximate phase shift?

•Using your previous answers, how could you approximate your model by a single sine or cosine function?

4.Now suppose you fix the mass of the weight at 2 kilograms, and assume the conditions wehad in problem 2 (the weight starts out 0.5 meters above equilibrium with no initial push). Draw separate sketches of your model with k = 2, k = 8, and k = 18 over the interval [0,4]. What do you notice about the graphs’ behaviour as you increase the value of k? How would you interpret this physically?

5.Now suppose you fix the spring constant at 20 Newtons per meter, and assume the conditionswe had in problem 2 (the weight starts out 0.5 meters above equilibrium with no initial push). Draw separate sketches of your model with m = 10, m = 20, and m = 40 over the interval [0,4]. What do you notice about the graphs’ behaviour as you increase the value of m? How would you interpret this physically?

Explanation / Answer

We use Hooke's law directly

F =-K x K = - F/x = 20/2 = 10 N/m

Second part

X (t) = Cos (K t m r) u + v0 r m K

For the amplitude of the movement

. t = 0

X (0) = uo + v0 rmK = 0.5

As there is no initial impulse v = 0

X (0) = u0 = 0.5 m

For drawing, were graph x(t) for various angles, for example

Angle X (t)

rad

0 0 0.5

45 /4 0.5 Cos 45 ° = 0.35 m

90 /2 0

135 3/4 - 0.35

180 - 0.5

The period the time that it takes to give a full swing or go from 0 to 2

Cos (K T m r) = cos 2

. r K m T = 2 T = 2 / (r m K) this equation is dimensionally incorrect, should be r t sqrt(k/m)

you do not indicate what r we suppose to be the range of motion

T = 2 / (0.5 10. 2) = /5 s

As there is no initial boost phase must be zero

X (0) = A Cos 0 + 0 = A

X (t) = A Cos (rm Kt + phase)

Third party

. v0 = 2 m/s

Now rewrite the equation.

X (t) = u Cos (r 10 0.5 t) + 2 10 0.5 r

For the chart do the same from the front

We are looking for the amplitude

Angle zero implies time zero

u Cos 0 + 10 r = 0.5

r should be a fact, or give something to calculate it

. u + 10 r = 0.5 r = (0.5-u) / 10

We need another equation angle /2 u Cos () + 10 r =-0.5 r = - (0.5 + u) 10

We add these two equations 20r =0 inconsistent

Angle X (t)

Rad

0 0 0.5

45 /4 u Cos /4 + 10r

90 /2 10r

The phase is t = 0 s

X (0) = u 0 Cos + 10r = u Cos Phase Phase = Cos-1 (10r)

Fourth part

Repeats changing K

Fifth part

. m = 10

X (t) = Cos (K t m r) u + v0 r m K

X (t) = u Cos (r 20 10 t) + 0

Amplitude

. t = 0 X (0) = u = 0.5

aquation X (t) = 0.5 (r 200t) I assume that r was a mistake or r = 1

Graphic

Angle X (t) t

Rad m = 10 m = 20

0 0 0.5 0 0

45 /4 0.35 /4 1/200=/4 0.005 /4 1/400=/4 0.0025

90 /2 0 /2 1/200 /2 1/400

We can see that you time is increasing as increasing m, by aso is better to see it with angles that remains constant

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