# Physics: Energy of a Spring--Harmonic Motion?

+1 vote

Physics: Energy of a Spring--Harmonic Motion?

An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.

What is the system's potential energy when its kinetic energy is equal to (3/4)E? (multiple choice question)

And the choices are:

kA^2

(kA^2)/2

(kA^2)/4

(kA^2)/8

According to the formula in the book, E=(1/2)kA^2, so wouldn't (3/4)E=(3/8)kA^2? But it's not one of the choices.

Please help me. Thanks.

Update:

Thanks BluePhysicsTea for Part I, here's the other part of the problem:

What is the object's velocity when its potential energy is (3/4)E?

+-(sqrt(k/m))A

+-(sqrt(k/m))(sqrt(2/3))A

+-(sqrt(k/m))(A/sqrt(3))

+-(sqrt(k/m))(A/sqrt(6))

## 3 Answers

0 votes
by
selected by

Best answer

Total energy= PEmax on the spring = KEmax of the oject

Total Energy= KE+PE at v or x not maximum.

Using the second line

E= 3/4 E + PE

PE = 1/4 E

PEmax =E = 1/2 kA^2

PE=E/4 = (1/2)kA^2 / 4= kA^2/8

take care

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.
0 votes
by

For the second part: First, none of those answers lines up with what is being asked, so none are correct. That said, if the question were asking what the velocity of the object was when potential energy is 2/3E, we could solve it as follows:

Total Energy E = PE + KE. So, if the object has a PE that is 2/3E, then it must have a KE = 1/3E.

As in the earlier part, Total Energy E is also 1/2(kA^2). We can set up two equations for KE.

KE=1/2mv^2

KE=1/3*(1/2*(kA^2))=1/6(kA^2)

We can then set those equations equal to each other and solve for v.

1/2(mv^2)=1/6(kA^2)

3(mv^2)=(kA^2)

(3m)v^2=(kA^2)

v^2=(kA^2)/(3m)

v^2=(k/m)*((A^2)/3)

v=+-(sqrt(k/m)*sqt((A^2)/3))=+-(sqrt(k/m)*(A/sqrt(3)))

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.
0 votes
by

"even as a nil.20-kg block is suspended from a vertically miraculous spring, it stretches the spring from its unique length of 0.040 m to 0.0.5 m." That tells you the spring consistent. you recognize the rigidity and the corresponding stretch. "The block is then pulled so as that the spring stretches to a finished length of 0.15 m." That tells you the total skill saved contained in the spring, (a million/2)kx^2 "what's the speed of the block each and every time the spring is 5.0 cm lengthy?" Calculate the flair skill at this fee of x. some thing of the skill is kinetic skill of the block.

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.

0 votes
1 answer
0 votes
3 answers
0 votes
2 answers
0 votes
1 answer
+1 vote
3 answers
0 votes
2 answers
0 votes
1 answer
0 votes
2 answers
0 votes
2 answers
0 votes
1 answer