Question

The distance, d, in inches of a weight attached to a spring from its equilibrium as a function in time, t, in seconds can be
modeled by the graph below. Which equation is represented in the graph below?​

Answer 1

Answer:

Please check the attachment

Step-by-step explanation:

The distance, d, of weight attached to the spring from its equilibrium is shown against θ is given in the attachment. You can converit into time, by using θ = ωT. And I have plotted the graph of d against θ  or ωT as ω is constant. As a result we have the desired graph.

Answer 2

The distance is root(2) times root(1-cos\theta)

The distance, d, of weight attached to the spring from its equilibrium is shown against θ is given in the attachment.

What is the formula for angular momentum?

θ = ωT.

ω=angular momentum

T=time

θ =angle

You can convert it into time, by using θ = ωT.

And I have plotted the graph of d against θ  or ωT as ω is constant.

As a result, we have the desired graph.

$d^2=(a -acos\theta)^2+sin\theta^2\\\\d=\sqrt{(a -acos\theta)^2+sin\theta^2} \\d=a(2-2cos\theta)\\d=\sqrt{2} a\sqrt{1-cos\theta}$

Let θ =0,30,60,90,180...a=1

$d=\sqrt{2} \sqrt{1-cos\theta} \\d=\sqrt{2} \sqrt{1-cos(0)}\\\\d=1.414$

Similarly, find for all angle

find the angle and draw a graph of the points.

Therefore we get the distance is root(2) times root(1-cos\theta).

To learn more about the distance visit:

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