Solving Displacement of Mass on Spring at v=1m/s

In summary: But remember to solve for x, you need to use the x(t) equation which is x(t) = Acos(ωt+φ). In summary, the question asks for the displacement of a 0.10kg mass on a spring with k=0.10 N/m when its velocity is 1.0m/s, given that it had a velocity of 2.0m/s at t=0s. Using the equations x(t)=Acos(ωt+φ) and v(t)=-wAsin(ωt+φ), we can solve for the displacement by setting v=1.0m/s and solving for x. Alternatively, we can use the total energy equation Etotal=1/
  • #1
Glorzifen
25
0

Homework Statement


A 0.10kg mass is on a spring with k = 0.10 N/m. At time t=0s it crosses x=0m with a velocity of 2.0m/s. What is its displacement when v=1.0m/s?

Homework Equations


x(t) = Acos(wt + [tex]\varphi[/tex])
v(t) = -wAsin(wt + [tex]\varphi[/tex])
w = [tex]\sqrt{k/m}[/tex]

The Attempt at a Solution


According to the question...shouldn't v(0) = 2? When a plug those numbers into the v(t) equation I get: v(0) = -wAsin(0) = 0...in other words, v(0) = 0...what am I missing here? I have an idea of where to go from here, but if someone could clarify how I'm supposed to start, that would be great. Thanks for your time.
 
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  • #2


Glorzifen said:

The Attempt at a Solution


According to the question...shouldn't v(0) = 2? When a plug those numbers into the v(t) equation I get: v(0) = -wAsin(0) = 0...in other words, v(0) = 0...what am I missing here? I have an idea of where to go from here, but if someone could clarify how I'm supposed to start, that would be great. Thanks for your time.

Remember you have x=Acos(ωt+φ) so x(0) = Acosφ

v=Aωsin(ωt+φ), v(0) does not give you Aωsin(0), remember there is still φ which is unaffected by t being equal to zero.
 
  • #3


rock.freak667 said:
Remember you have x=Acos(ωt+φ) so x(0) = Acosφ

v=Aωsin(ωt+φ), v(0) does not give you Aωsin(0), remember there is still φ which is unaffected by t being equal to zero.

Okay - not sure if this route even needs to be undertaken at all though...does this work?:

Etotal = [tex]\frac{1}{2}[/tex]kx2 + [tex]\frac{1}{2}[/tex]mv2

Etotal = 0 + [tex]\frac{1}{2}[/tex]0.1(2)2

Etotal = 0.2

0.2 = [tex]\frac{1}{2}[/tex](0.1)x2 + [tex]\frac{1}{2}[/tex]0.1(1)2

x = 1.7

Does that work?
 
  • #4


Yes that works just as well.
 

1. How do you calculate the displacement of mass on a spring at a velocity of 1m/s?

To calculate the displacement of mass on a spring at a velocity of 1m/s, you will need to use the equation x=mv/k, where x represents displacement, m represents mass, v represents velocity, and k represents the spring constant. Plug in the known values and solve for the displacement.

2. What is the spring constant and how does it affect the displacement of mass on a spring?

The spring constant, represented by the variable k, is a measure of how stiff or flexible a spring is. It is calculated by dividing the applied force by the resulting displacement. The higher the spring constant, the stiffer the spring and the less it will stretch when a mass is attached. This means that a higher spring constant will result in a smaller displacement for the same mass and velocity.

3. Can the displacement of mass on a spring be negative?

Yes, the displacement of mass on a spring can be negative. This means that the mass has moved in the opposite direction of the applied force. It is important to consider the direction of the displacement when solving problems involving springs.

4. What are the units for displacement of mass on a spring?

The units for displacement of mass on a spring are meters (m). This is because displacement is a measure of distance and is typically represented by the symbol x. In the SI system, the standard unit for distance is meters.

5. How does the velocity of the mass affect the displacement on a spring?

The velocity of the mass affects the displacement on a spring by directly impacting the amount of force applied to the spring. According to Newton's Second Law of Motion, force is equal to mass multiplied by acceleration. In this case, the acceleration is represented by the velocity. So, a higher velocity will result in a greater force and therefore a larger displacement.

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