spring constant

[Muzette]

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TBR Physics Ch. 5 #22 and #23 is are relevant to this figure.
#22 If the experiment were repeated with a heavier mass and the same initial displacement, then how would the results vary?
ANS: The mass would have greater momentum at equilibrium

I don't under stand why momentum would vary. The displacement is the same, which means potential energy is the same, which means kinetic energy is the same for varying masses. So if 1/2mv^2 does not vary, why does mv^2 vary?

#23 When two springs of different materials are used in the experiment, what should be observed?
ANS: The mass oscillate in a symmetric fashion

What does it really imply when the mass is oscillating in a symmetric fashion? Does it mean that the displacement is symmetrical? Please help!!

 

[Jepstein30]

TBR Physics Ch. 5 #22 and #23 is are relevant to this figure.
#22 If the experiment were repeated with a heavier mass and the same initial displacement, then how would the results vary?
ANS: The mass would have greater momentum at equilibrium

I don't under stand why momentum would vary. The displacement is the same, which means potential energy is the same, which means kinetic energy is the same for varying masses. So if 1/2mv^2 does not vary, why does mv^2 vary?

F = -kx
PE(spring) (kx^2)/2
KE = (mv^2/2)

Here, we are converting between work done by the spring and SPRING potential energy, not gravitational potential energy (which is constant because the spring is horizontal).

So since you are only changing the mass of the object, you aren't affecting the F(spring) or PE(spring). But, you ARE increasing the KE and momentum.

when dealing with Gravitational potential energy, mass cancels out:
mgh = 1/2mv^2
gh = 1/2v^2

but it doesn't here:
kx^2/2 = mv^2/2
kx^2 = mv^2

#23 When two springs of different materials are used in the experiment, what should be observed?
ANS: The mass oscillate in a symmetric fashion

What does it really imply when the mass is oscillating in a symmetric fashion? Does it mean that the displacement is symmetrical? Please help!!

I don't get what you (or the question really) is asking. What are the other answer choices? Simple harmonic motion will ALWAYS be symmetrical (someone correct me if I'm wrong but I've never seen a non-symmetrical simple system). If you have dampening, I guess it won't be symmetric because each time the amplitude would be decreasing.

are you not supposed to consider friction?

 

[milski]

Keep in mind, momentum is mv, not mv². If the mass increases from m to Am, the speed at equilibrium will decrease to v/sqrt(A) and mass will be mA. Then the momentum is mA*v/sqrt(A)=mv.sqrt(A) or an increase of sqrt(A).

 

[Muzette]

Hey thank you for your reply! For the first question, the explanation says that:

If a heavier mass were used with the original displacement, then the system would still start with the same initial potential energy (PEsprjng = 1/2^kx2). This results in the same kinetic energy as the object passes through its equilibrium position. Because kinetic energy is unchanged despite the greater mass, it must be moving more slowly with a heavier mass (to have the same value for 1/2mv^2). Because the kinetic energy is the same for both the lighter and heavier masses,we can say they each have the same value for mv^2. So, the mass is increased by a greater factor than the velocity is decreased, because an increase in mass is exactly balanced by a decrease in v^2.Because mass is more important than velocity, the heavier object has a greater momentum as it passes through its equilibrium point than the lighter object.

Sorry for bombarding you with a bunch of text, but it seems contrary to what you said (although your explanation makes more sense to me). It's saying that the KE does not change but momentum does.

For the second question, I was wondering if the springs on either side had different k values, it would affect the symmetric oscillation of the system.
The exact question is this:
When two springs of different materials are used in the experiment, what should be observed?
A. The mass oscillates in a symmetric fashion
B. A greater displacement is observed on the side with the spring having the smaller spring constant.
C. The potential energy is greater when the object is at rest on the side with the greater k than when the object is at rest on the side with the smaller k.
D. The object reaches a greater maximum of speed in one direction than the other.
The answer is A.
Please continue to help me with these questions. Thank you!!

 

[BerkReviewTeach]

Sorry for bombarding you with a bunch of text, but it seems contrary to what you said (although your explanation makes more sense to me). It's saying that the KE does not change but momentum does.

Consider any two objects that have different masses, but the same kinetic energy. To have the same ¹/₂mv², the ratio of the masses for the two objects must equal the inverse of the ratio of the v²s. So if one object is twice as heavy as the other, then its v² is half as large, which ultimately means that its v is ¹/_{(root 2)} as large.

So m_larger = 2m_smaller and v_smaller = (root 2)v_larger.

Momentum for the larger ends up being equal to 2m_smaller x ¹/_{(root 2)}v_smaller = (root 2)m_smallerv_smaller, which makes it larger by a factor of root 2. Hopefully the math helps. The point of all of this is to say that having the same KE but a different mass means that the objects must have different momenta.

For the second question, I was wondering if the springs on either side had different k values, it would affect the symmetric oscillation of the system.
The exact question is this:
When two springs of different materials are used in the experiment, what should be observed?
A. The mass oscillates in a symmetric fashion
B. A greater displacement is observed on the side with the spring having the smaller spring constant.
C. The potential energy is greater when the object is at rest on the side with the greater k than when the object is at rest on the side with the smaller k.
D. The object reaches a greater maximum of speed in one direction than the other.
The answer is A.
Please continue to help me with these questions. Thank you!!

Perhaps it is best to think about the object at different positions. Let's say the spring on its left is stiffer than the spring on its right (k_left > k_right). If you push the object to the left, you are working harder to compress the left spring than to stretch the right spring, so you are adding a large work value (for the left side) to a small work value (for the right side) to get the total amount of work needed to move the object. So what happens if you move the object the other way? If you push the object to the right, you are working harder to stretch the left spring than to compress the right spring, so you are once again adding a large work value (for the left side) to a small work value (for the right side) to get the total amount of work needed to move the object.

Whether you displace the object to the left or right, you are doing the same work, so it is symmetric in terms of left versus right.

Math wise, deltaPE = ¹/₂k_leftx² + ¹/₂k_rightx² no matter which way you displace the object from equilibrium, so its symmetric from a left-right perspective.

 

[sazerac]

Here is an alternative way to think about the first question.

Kinetic energy is force x distance. If you make the mass heavier, are you changing the force exerted by the spring? No. Are you changing the distance over which the spring acts? No. So the kinetic energy will be the same.

Momentum is force x time. If you make the mass heavier the force won't change, but the time sure will increase. Takes a lot longer to push a bowling ball around than a ping pong ball, if forces are the same! Same force, more time, therefore the momentum will increase too.

Being able to quickly solve comparisons like these on the MCAT without resorting to square roots and stuff will buy you a lot of time to work on other problems.

 

[gjh]

Perhaps it is best to think about the object at different positions. Let's say the spring on its left is stiffer than the spring on its right (k_left > k_right). If you push the object to the left, you are working harder to compress the left spring than to stretch the right spring, so you are adding a large work value (for the left side) to a small work value (for the right side) to get the total amount of work needed to move the object. So what happens if you move the object the other way? If you push the object to the right, you are working harder to stretch the left spring than to compress the right spring, so you are once again adding a large work value (for the left side) to a small work value (for the right side) to get the total amount of work needed to move the object.

Whether you displace the object to the left or right, you are doing the same work, so it is symmetric in terms of left versus right.

Math wise, deltaPE = ¹/₂k_leftx² + ¹/₂k_rightx² no matter which way you displace the object from equilibrium, so its symmetric from a left-right perspective.

I'm still stuck on this explanation -- I understand that the restoring force acting on the mass is the same on both sides when compressed, but why is the displacement from equilibrium the same in either direction? Are we assuming the k values are the same for the two springs? Wouldn't the k values be different for the two springs if they were made of different material? TBR says in their rationale that "The displacement from equilibrium is equal on both sides, because the restoring force is identical whether the spring has been elongated or compressed." F=-kx: If we know that the restoring F is the same for both springs, how can we determine x if k is unknown as well?

 

[The_Sunny_Doc]

I'm still stuck on this explanation -- I understand that the restoring force acting on the mass is the same on both sides when compressed, but why is the displacement from equilibrium the same in either direction? Are we assuming the k values are the same for the two springs? Wouldn't the k values be different for the two springs if they were made of different material? TBR says in their rationale that "The displacement from equilibrium is equal on both sides, because the restoring force is identical whether the spring has been elongated or compressed." F=-kx: If we know that the restoring F is the same for both springs, how can we determine x if k is unknown as well?

Remember that when you move the mass the one side, you're squishing one spring just as much as you're stretching the other spring. Let's say that the spring on the left side of the block is more stiff. When you squish the mass to left, away from equilibrium, the left spring will snap back to the right "harder" than the right spring will pull right. As the mass travels to the right, it faces less resistance to squish the weaker spring on the right, so you'd think the mass would move farther to the right from equilibrium. However, the block is yanked left by that stiffer spring to make up for for the weaker push back of the spring on the right. It's a tug of war with each spring contributing their part to keep the mass oscillating with the same cycles per second.

This pic might help.