TBR physics Section 4 Question 27

[SeeEll]

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We're dealing with simple harmonic motion. My brain just might be fried but I am not understanding how we get to this equation as shown in the solutions.

The question asks:

"At what position x will the speed of the mass be half its maximum speed, Vmax"?

They've shown a method to solve it by setting the total energy equal to the kinetic energy. I have some questions here.

1) Why is it that we can set both KE and PE to 1/2 MVmax^2 when dealing with a maximum kinetic energy?

2) They solve it out by setting up the problem in the following way:

1/2MVmax^2 = 1/2MV^2 + 1/2KX^2

1/2KX^2 = 1/2MVmax^2 - 1/4(1/2MVmax^2)

This is probably super elementary. My physics has never been that great. Can anyone explain why that 1/4th goes in there?

 

[loveoforganic]

I don't think I understand your first question.

The answer to your second question - At 1/2 the maximum velocity (1/2 Vmax^2), you will have 1/4 the total energy present in the system (made up of kinetic and potential energy). This is because, by the relation KE = 1/2 mv^2, if you half the velocity (1/2 Vmax), you quarter the kinetic energy.

 

[SeeEll]

I don't know why I keep having trouble relating these things in physics. Forget the first part. Thank you for the explanation.

 

[Rabolisk]

I think I understand your first question. KEmax is equal to the total energy at any time because in SHM, total energy is constant. Since total energy is te sum of KE and PE, you can equate KEmax to KE + PE.

 

[bbottle]

You can also set it to PEmax (1/2KXmax^2)...the key is that when KE is at its maximum, PE is 0 and vice-versa.

KE + PE = KEmax + 0 = 0 + PEmax

 

[SeeEll]

Thanks to the both of you.