Calculating total mechanical energy of block on spring

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  • #1
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Homework Statement


A 5.0 kg block hangs from a spring with spring constant 2000N/m. The block is pulled down 5 cm from equilibrium position and given an initial velocity of 1.0 m/s back toward equilibrium. What are the

a.) frequency
b.) amplitude
c.) total mechanical energy of the motion


Homework Equations


oscillatory motion x(t) = Acos(ωt + ø0)
f = ω/(2∏)
ω = √(k/m)


The Attempt at a Solution



frequency found by ω = √(k/m)
ω = √(2000/5) = 20
f = 20/(2∏) = 10/∏
which I believe was correct.

The only way to find the amplitude is to relate it to the velocity, correct? Here I can't simply use .5kA2 = .5m(Vmax)2 because at Vmax there should also be some gravitational potential. Used the motion equation and it's derivative:

x(t) = Acos(ωt + ø0)
v(t) = -ωAsin(ωt + ø0)

x(0) = .05 = A cos(ø0)
=>ø0 = cos-1(.05/A)

edit: here I made an error v(0) should be -20Asin(ø0)
v(0) = -1 = -20A/sin(ø0)
substituting...
-1 = -20Asin(cos-1(.05/A))
=>sin-1(1/(20A)) = cos-1(.05/A)

which is the same thing as saying (.05)2 + (.05)2 = A2 right? If so, A = .0707, which I can't verify at the moment but seems reasonable.

Where I get stuck is here:
The energy should be entirely kinetic at the bottom of the stretch given by .5k(A + Δx)2 where Δx was the initial stretch of the spring caused by gravity before it was pulled. So I figured I could set it equal to the initial condition when the spring is displaced .05 m from equilibrium and the initial velocity is 1.0 m/s:
.5k(A + Δx)2 = .5k(.05 + Δx)2 + mg(A-.05) + .5mV02
Then I solved for Δx, in order to plug it back into .5k(A + Δx)2 and I won't bore you with all that here but I've done it twice now and Δx comes out -.1549, which doesn't seem right. Plugging it into the equation negative or positive comes out as 26.6 and 223.6 respectively, and I believe the answer was supposed to be 5 as total mechanical energy.
After doing it again with the correct amplitude I got 8.94, a lot closer to 5.0 but still off...

If anyone sees my errors here I'd be super grateful!
 
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Answers and Replies

  • #2
rude man
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Homework Statement


A 5.0 kg block hangs from a spring with spring constant 2000N/m. The block is pulled down 5 cm from equilibrium position and given an initial velocity of 1.0 m/s back toward equilibrium. What are the

a.) frequency
b.) amplitude
c.) total mechanical energy of the motion


Homework Equations


oscillatory motion x(t) = Acos(ωt + ø0)
f = ω/(2∏)
ω = √(k/m)


The Attempt at a Solution



frequency found by ω = √(k/m)
ω = √(2000/5) = 20
f = 20/(2∏) = 10/∏
which I believe was correct.

The only way to find the amplitude is to relate it to the velocity, correct? Here I can't simply use .5kA2 = .5m(Vmax)2 because at Vmax there should also be some gravitational potential. Used the motion equation and it's derivative:

x(t) = Acos(ωt + ø0)
v(t) = -ωAsin(ωt + ø0)

x(0) = .05 = A cos(ø0)
=>ø0 = cos-1(.05/A)

edit: here I made an error v(0) should be -20Asin(ø0)
v(0) = -1 = -20A/sin(ø0)
substituting...
-1 = -20Asin(cos-1(.05/A))
=>sin-1(1/(20A)) = cos-1(.05/A)

which is the same thing as saying (.05)2 + (.05)2 = A2 right? If so, A = .0707, which I can't verify at the moment but seems reasonable.
Right so far.

Where I get stuck is here:
The energy should be entirely kinetic at the bottom of the stretch
Nope. The energy at the bottom is the kinetic energy plus the energy stored in the spring.
 
  • #3
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1
Whoops, what I meant to say was that it should be all spring potential at the bottom of the stretch... there shouldn't be any kinetic because we're talking about the moment the block changes direction from down to up?
 
  • #4
78
12
Also remember that simply because it's suspended above the Earth some distance, it starts out with some potential energy. Which you will never see unless the spring breaks. :)
 
  • #5
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1
Also remember that simply because it's suspended above the Earth some distance, it starts out with some potential energy. Which you will never see unless the spring breaks. :)
Hmm, I set the gravitational potential equal to 0 at the full extension of the spring downward. That would work right?
 
  • #6
rude man
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Whoops, what I meant to say was that it should be all spring potential at the bottom of the stretch... there shouldn't be any kinetic because we're talking about the moment the block changes direction from down to up?
No, the problem states that the initial velocity is 1 m/s so the initial k.e. must also be non-zero.

Gravitational potential energy enters the picture but for your purposes you can call the zero p.e. point the initial point, i.e when the mass is initially stretched 5 cm beyond its natural stretch. That way you can ignore it in answering your questions.
 
  • #7
37
1
No, the problem states that the initial velocity is 1 m/s so the initial k.e. must also be non-zero.

Gravitational potential energy enters the picture but for your purposes you can call the zero p.e. point the initial point, i.e when the mass is initially stretched 5 cm beyond its natural stretch. That way you can ignore it in answering your questions.
Right, but at some point when the spring comes back down it will reach it's full amplitude and all the energy would be potential then (some extension beyond the initial condition x= .05) So I wanted to set that equal to the initial condition, then solve the entire expression for the unknown variable. But to do that I would have to set the potential to zero at the bottom of the stretch or I would have negative potential at the full extension and I'm not sure what that means.

Is there another way to do it?
 
  • #8
rude man
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Right, but at some point when the spring comes back down it will reach it's full amplitude and all the energy would be potential then (some extension beyond the initial condition x= .05) So I wanted to set that equal to the initial condition, then solve the entire expression for the unknown variable. But to do that I would have to set the potential to zero at the bottom of the stretch or I would have negative potential at the full extension and I'm not sure what that means.

Is there another way to do it?
Why not just go with the initial conditions? That's all you have to do. The total energy is conserved after that; it never changes. So all you have is the initial k.e. plus the initial spring-stored energy.

There's a bit of definiton involved here: the total mechanical energy is the sum of kinetic and stored spring energy assuming the initial spring-stored energy does not include the spring-stored energy before the mass is pulled by the additional 5 cm. The argument here is that if we start at the point where the mass is held such that the spring is initially relaxed, that represents zero total mechanical energy. Which is the same as what I defied above since the drop in p.e. when the mass is allowed to dangle freely = the energy stored in the spring: mg delta x = 1/2 kx^2.
 
  • #9
rude man
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Also: the velocity at the bottom of each swing is not zero!
Look at you equations for x(t) and v(t) and you will see.
(The reason is the initial velocity of 1 m/s imparted to the mass).
 

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