Springs: speed and acceleration

In summary: It's much appreciated.In summary, the object vibrates with an amplitude of 11 cm and a speed of -0.145 cm/second.
  • #1
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Homework Statement


A 0.450 kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.0 cm.
Calculate the maximum value (magnitude) of its speed and acceleration.[/B]

The Attempt at a Solution



I found both w and t.

[tex] \omega = 4.216[/tex]
[tex] t = 0.145[/tex]

To get t I solved this equation for t.

[tex] x(t) = Acos(\omega t + \phi)[/tex]
[tex]t = \frac{1}{4.216}cos^-1(\frac{x}{11})[/tex]


With that value of t, I solved for v. I got a negative number which is incorrect. However, if at the start I use sine instead of cosine for position i get a different value for t, and get the same answer for v, except positive, which is correct. But why would I use sine for position rather then cosine, and how would I know? Also why would it be positive or negative?
 
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  • #2
Why are you solving for 't'. What t? Just put x(t)=A*cos(omega*t). Then v(t)=x'(t) and a(t)=x''(t). Work those out. Now to the find the maxes you don't even have to worry about the trig function. The max of either sin or cos is 1.
 
  • #3
I am sorry... I left off part of the question...

(b) Calculate the speed and acceleration when the object is 9.00 cm from the equilibrium position.

That is why I was finding t
 
  • #4
Sheneron said:
I am sorry... I left off part of the question...

(b) Calculate the speed and acceleration when the object is 9.00 cm from the equilibrium position.

That is why I was finding t

Ah. In that case the problem doesn't give you enough information to determine a sign for v or a. The object could be 9 cm from equilibrium and either moving away from or towards equilibrium. They seem to just want the magnitude of v and a. Not the sign. You can use either sin or cos. They only differ by a change the phase angle (phi).
 
  • #5
As an alternative, you still are not required to find the time for part b. You can use conservation of energy to find the speed and a force diagram for the magnitude of the acceleration.
 
  • #6
Dick said:
Ah. In that case the problem doesn't give you enough information to determine a sign for v or a. The object could be 9 cm from equilibrium and either moving away from or towards equilibrium. They seem to just want the magnitude of v and a. Not the sign. You can use either sin or cos. They only differ by a change the phase angle (phi).

Yes that's what thought, but I submitted Acceleration with a postive sign, and it was counted wrong, but with a negative sign correct.

alphysicist said:
As an alternative, you still are not required to find the time for part b. You can use conservation of energy to find the speed and a force diagram for the magnitude of the acceleration.

I'll try this method thank you.
 
  • #7
Also, the thing I didn't understand was... I have a solutions manual and this problem was in there. I checked the book and it started with x(t) = Asin(wt + phi), rather then cosine. Which gave the correct sign for direction. But how would you know whether to use cosine or sine? Generally you use cosine don't you?
 
  • #8
That's silly. If they had wanted you to get that specific setup, they should have told you. sin(x+pi/2)=cos(x). There is no difference except for the phase. Even if you use sin, there are 4 times every cycle when the object is 9cm away from equilibrium - and they have all possible combinations of signs for a and v.
 
  • #9
Well, Thanks for your help.
 

What is the relationship between speed and acceleration in a spring?

The speed and acceleration of a spring are directly proportional to each other. This means that as the speed of the spring increases, so does its acceleration, and vice versa.

How do you calculate the speed of a spring?

The speed of a spring can be calculated using the equation v = ωA, where v is the speed, ω is the angular frequency, and A is the amplitude (maximum displacement) of the spring.

What factors affect the acceleration of a spring?

The acceleration of a spring is affected by factors such as the mass of the object attached to the spring, the spring constant (a measure of the stiffness of the spring), and any external forces acting on the spring.

Can the speed of a spring exceed the speed of light?

No, the speed of a spring is limited by the speed of sound, which is much slower than the speed of light. Therefore, it is impossible for the speed of a spring to exceed the speed of light.

How does the speed and acceleration of a spring change with different masses attached?

The speed and acceleration of a spring will change with different masses attached. As the mass increases, the speed and acceleration of the spring will decrease, and vice versa. This is because the mass affects the spring constant and the force applied to the spring, which in turn affects its speed and acceleration.

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