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

AI Thread Summary
A 0.10kg mass on a spring with a spring constant of 0.10 N/m is analyzed for its displacement when its velocity is 1.0 m/s, starting from an initial velocity of 2.0 m/s. The user initially struggles with the equations of motion, particularly the relationship between velocity and displacement at t=0. Clarification is provided that the phase constant φ affects the initial conditions, leading to the correct application of the equations. The total energy method is then employed, confirming that the displacement when the velocity is 1.0 m/s is approximately 1.7 meters. This approach effectively resolves the problem using energy conservation principles.
Glorzifen
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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 + \varphi)
v(t) = -wAsin(wt + \varphi)
w = \sqrt{k/m}

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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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.
 


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 = \frac{1}{2}kx2 + \frac{1}{2}mv2

Etotal = 0 + \frac{1}{2}0.1(2)2

Etotal = 0.2

0.2 = \frac{1}{2}(0.1)x2 + \frac{1}{2}0.1(1)2

x = 1.7

Does that work?
 


Yes that works just as well.
 
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