- #1

- 11

- 0

I just have to use KE=1/2mv^2 right?

do I have to involve the amplitude?

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- Thread starter nicoleb14
- Start date

- #1

- 11

- 0

I just have to use KE=1/2mv^2 right?

do I have to involve the amplitude?

- #2

- 998

- 15

In SHM the KE is continually varying. So you are asking for the MAX KE, right?

- #3

- 11

- 0

I just need the energy of motion, so i guess that means KE max?

- #4

- 1,506

- 18

You are correct to use KE=1/2mv^2 but you need an expression for vmax in simple harmonic motion. You will need to find the spring constant (stiffness) and use this to find the max velocity.

Do you know these equations?

- #5

- 998

- 15

What is the max linear velocity in SHM?

- #6

- 11

- 0

Vmax=Aw

- #7

- 998

- 15

Then you can find max ke = (1/2)mv^2 and use v = Aw

- #8

- 11

- 0

thank you sooo muuuch!!!!! :D

- #9

PeterO

Homework Helper

- 2,435

- 62

I just have to use KE=1/2mv^2 right?

do I have to involve the amplitude?

Is this mass hanging from the spring, or is this whole situation taking place horizontally??

- #10

- 11

- 0

the problem doesnt say

- #11

PeterO

Homework Helper

- 2,435

- 62

the problem doesnt say

Assuming it is vertical, there is a constant interchange of elastic energy in the spring, gravitational potential energy and kinetic energy.

The total of them at any time will be constant.

That might be what is meant by the energy of the motion.

Even if the situation is horizontal [so no change in gravitational potential energy] there is a steady interchange between elastic potential energy and kinetic energy.

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