Velocity for Spring Constant with Amplitude

[MissPenguins]

Homework Statement

A 79.8 g mass is attached to a horizontal
spring with a spring constant of 2.66 N/m
and released from rest with an amplitude of
39.5 cm.
What is the velocity of the mass when it
is halfway to the equilibrium position if the
surface is frictionless?
Answer in units of m/s.

Homework Equations

v = $$\pm$$$$\sqrt{}$$k/m(A²-x²

The Attempt at a Solution

So I used the above equation, and got 2.280533. The answer is wrong. Did I use the correct equation? Thanks.

 

[thebigstar25]

the answer required is in m/s ..

I tried myself to use the equation you wrote and i obtained different answer than the one you got .. so I would suggest you to look more carefully at the units ..

convert gram to kilogram .. cm to m ..

 

[MissPenguins]

the answer required is in m/s ..

I tried myself to use the equation you wrote and i obtained different answer than the one you got .. so I would suggest you to look more carefully at the units ..

convert gram to kilogram .. cm to m ..

I converted everything, and got the same answer.

 

[thebigstar25]

can you please write the steps you followed in detials so i can check where you did wrong..

 

[MissPenguins]

$$\sqrt{}(2.66N/m / 0.0798 kg)(0.395)^2$$ = 2.2805

Thanks.

 

[thebigstar25]

the first thing you missed is x? .. you have in the equation you supplied us with a term for x^2 .. I can not see it in your solution ..

note: your question is asking you to find the velocity when it reaches half way to its equilibrium position .. what you calculated is (wA) which I believe is the maximum velocity..

 

[MissPenguins]

the first thing you missed is x? .. you have in the equation you supplied us with a term for x^2 .. I can not see it in your solution ..

note: your question is asking you to find the velocity when it reaches half way to its equilibrium position .. what you calculated is (wA) which I believe is the maximum velocity..

No, it's not x². It is the square root of the whole thing, and inside the square root, I square the 0.395. And I believe it's not the maximum velocity because v_maxhas a totally different equation. Thanks.

 

[|<ings]

And I believe it's not the maximum velocity because v_maxhas a totally different equation. Thanks.

MissPenguins, please note that if the equation is valid, the v_max would be given by setting x equal to 0. Also, I'm fairly sure that bigstar is correct: you forgot to subtract x² from A².

 

[thebigstar25]

so are you saying that you got for (A^2 - x^2) term a value of 0.395^2 ? .. this still different from what i got?? ..

What value you substitute for x in the equation?

 

[MissPenguins]

MissPenguins, please note that if the equation is valid, the v_max would be given by setting x equal to 0. Also, I'm fairly sure that bigstar is correct: you forgot to subtract x² from A².

Isn't x²=0 since x is not given?

 

[MissPenguins]

so are you saying that you got for (A^2 - x^2) term a value of 0.395^2 ? .. this still different from what i got?? ..

What value you substitute for x in the equation?

I thought x is 0 since it's not given. Am I wrong?

 

[thebigstar25]

noooooo .. Read the question carefully ..

it said find the velocity when it is halfway from the equilibrium .. If it said when it reaches the equilibrium then you put x=0 ..

Try again, and think more deeply about what value for x you should use ..

 

[|<ings]

I thought x is 0 since it's not given. Am I wrong?

Miss Penguins read bigstar's post and keep in mind that just because a value for a quantity is not given (in this case it is given, just not explicitly), you can't assume that the quantity's value is 0.

BTW, I don't have to be a mentor to give help, do I? (I'm new)

 

[thebigstar25]

Miss Penguins read bigstar's post and keep in mind that just because a value for a quantity is not given (in this case it is given, just not explicitly), you can't assume that the quantity's value is 0.

BTW, I don't have to be a mentor to give help, do I? (I'm new)

no you don't have to .. But you should follow the rules of this forum .. You are not allowed to solve the problem just few hints can be given...

 

[MissPenguins]

noooooo .. Read the question carefully ..

it said find the velocity when it is halfway from the equilibrium .. If it said when it reaches the equilibrium then you put x=0 ..

Try again, and think more deeply about what value for x you should use ..

When it says halfway, which is something divide by 2, correct?

 

[thebigstar25]

okay that is a good start .. Can you figure out what should this something be ..

Hint: try to make a simple drawing of the problem, things may get clearer for you regarding the choice of that ''something'' ..

 

[Bearbull24.5]

Would x be half the Amplitude? I have this same problem and only one chance left to get the correct answer.

 

[thebigstar25]

Would x be half the Amplitude? I have this same problem and only one chance left to get the correct answer.

You tell me, why do you think that x is half the amplitude?

 

[Bearbull24.5]

The amplitude is how far the spring moves in either direction from equilibrium. The question wants to know how fast it is moving when it is half the distance from the equilibrium which would be one half of the amplitude.

 

[thebigstar25]

The amplitude is how far the spring moves in either direction from equilibrium. The question wants to know how fast it is moving when it is half the distance from the equilibrium which would be one half of the amplitude.

Hmm, well .. You convinced me with your answer :) , let's see whether the last answer you give would give you the right solution ..

 

[Bearbull24.5]

Hmm, well .. You convinced me with your answer :) , let's see whether the last answer you give would give you the right solution ..

Got it right. Thanks

 

[thebigstar25]

:) you are welcome .. You had the right answer from the beginning I didnt do that much ..

 

[MissPenguins]

[SOLVED] Velocity for Spring Constant with Amplitude

:) you are welcome .. You had the right answer from the beginning I didnt do that much ..

yay, I got it right too. Thank you to you and everyone else who helped out. ;)

 

[MissPenguins]

[SOLVED] Velocity for Spring Constant with Amplitude

Homework Statement

A 79.8 g mass is attached to a horizontal
spring with a spring constant of 2.66 N/m
and released from rest with an amplitude of
39.5 cm.
What is the velocity of the mass when it
is halfway to the equilibrium position if the
surface is frictionless?
Answer in units of m/s.

Homework Equations

v = $$\pm$$$$\sqrt{}$$k/m(A²-x²

The Attempt at a Solution

So I used the above equation, and got 2.280533. The answer is wrong. Did I use the correct equation? Thanks.

Problem solved, thank you very much everyone. Your help is appreciated. ;)

 

[thebigstar25]

you are welcome .. Just next time read the question more carefully so you can extract hidden information :D ..