Question

A mass attached to a spring oscillates upward and downward. The length L of the spring after T seconds is givrn by the function L = 15 - 3.5cos(2piT). What is the length of the spring when it is at equilibrium (when the spring is still)? What is the length of the spring when it is the shortest? Longest?

Answer 1

I got shortest 11.5, longest 18.5...I am unsure about at rest but I think it's 11.5 unless im not understanding fully.

Answer 2

Finding the equilibrium point is as easy as plugging in a value of \(t\) that makes the cosine part 0.

The extremal points occur for values of \(t\) that maximize the cosine part.

Since cosine is bounded between -1 and 1, you can immediately see that the maximum length will be \(15-3.5(-1)\), and the minimum length will be \(15-3.5(1)\).

Answer 3

Could you explain the first part more? What is the equation I am solving for in the first part? \[L=15-3.5\cos(0)\]

Answer 4

That is the equilibrium length. You can think of it in another way. The equilibrium point occurs at the midpoint between maximum and minimum length, and 15 is between 18.5 and 13.5.

Answer 5

Sorry, *11.5

Answer 6

16!

Answer 7

halfway between 11.5 and 18.5 is?

Answer 8

not 16..

Answer 9

15.5 sorry i blacked out

Answer 10

I majored in English Language and Literature....I'm in the stratosphere.

Answer 11

\[\frac{11.5+18.5}{2}=\frac{30}{2}=15\not=16\not=15.5\]

Answer 12

is horizonal shift and phase shift the same thing?