Velocity for Spring Constant with Amplitude

In summary: When it says halfway, which is something divide by 2, correct?Okay, that is a good start. Can you figure out what should this something be?
  • #1
MissPenguins
58
0

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 = [tex]\pm[/tex][tex]\sqrt{}[/tex]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.
 
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  • #2
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 ..
 
  • #3
thebigstar25 said:
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.
 
  • #4
can you please write the steps you followed in detials so i can check where you did wrong..
 
  • #5
[tex]\sqrt{}(2.66N/m / 0.0798 kg)(0.395)^2[/tex] = 2.2805

Thanks.
 
  • #6
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..
 
  • #7
thebigstar25 said:
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 vmaxhas a totally different equation. Thanks.
 
  • #8
MissPenguins said:
And I believe it's not the maximum velocity because vmaxhas a totally different equation. Thanks.

MissPenguins, please note that if the equation is valid, the vmax 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.
 
Last edited:
  • #9
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?
 
  • #10
|<ings said:
MissPenguins, please note that if the equation is valid, the vmax 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?
 
  • #11
thebigstar25 said:
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?
 
  • #12
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 ..
 
  • #13
MissPenguins said:
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)
 
  • #14
|<ings said:
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...
 
  • #15
thebigstar25 said:
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?
 
  • #16
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'' ..
 
  • #17
Would x be half the Amplitude? I have this same problem and only one chance left to get the correct answer.
 
  • #18
Bearbull24.5 said:
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?
 
  • #19
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.
 
  • #20
Bearbull24.5 said:
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 ..
 
  • #21
thebigstar25 said:
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
 
  • #22
:) you are welcome .. You had the right answer from the beginning I didnt do that much ..
 
  • #23
[SOLVED] Velocity for Spring Constant with Amplitude

thebigstar25 said:
:) 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. ;)
 
  • #24
[SOLVED] Velocity for Spring Constant with Amplitude

MissPenguins said:

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 = [tex]\pm[/tex][tex]\sqrt{}[/tex]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. ;)
 
  • #25
you are welcome .. Just next time read the question more carefully so you can extract hidden information :D ..
 

1. What is the formula for calculating the velocity of a spring with a given spring constant and amplitude?

The formula for calculating the velocity of a spring with a given spring constant (k) and amplitude (A) is v = √(k/m) * A, where m is the mass of the object attached to the spring. This formula is derived from the equation for the potential energy of a spring, U = 1/2 * k * x^2, where x is the displacement of the spring from its equilibrium position.

2. How does the spring constant affect the velocity of a spring?

The spring constant (k) directly affects the velocity of a spring. A higher spring constant means that the spring is stiffer, and therefore will have a higher velocity for a given amplitude. Conversely, a lower spring constant will result in a lower velocity for the same amplitude.

3. Can the amplitude of a spring affect its velocity?

Yes, the amplitude of a spring does affect its velocity. The greater the amplitude, the higher the velocity will be. This is because a larger amplitude means that the spring is stretched or compressed more, resulting in a greater potential energy and therefore a higher velocity.

4. How does the mass of an object attached to a spring impact its velocity?

The mass of the object attached to a spring affects its velocity in two ways. Firstly, a heavier object will require a greater force to stretch or compress the spring, resulting in a lower velocity. Secondly, a heavier object will have a greater inertia, meaning it will resist changes in velocity and therefore have a lower velocity overall.

5. Is the velocity of a spring constant with amplitude always constant?

No, the velocity of a spring with a given spring constant and amplitude is not always constant. As the object attached to the spring moves, the velocity will change depending on the position of the object. At the equilibrium position, the velocity will be zero, and at the maximum amplitude, the velocity will be at its maximum. The velocity will also change over time as the spring oscillates back and forth.

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