Determine the frequency of the resulting motion

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Homework Statement


A 1.2-kg block rests on a frictionless surface and is attached to a horizontal spring of constant k = 27 N/m. The block is oscillating with amplitude 10 cm and phase constant phi = -pi/2. A block of mass 0.80 kg is moving from the right at 1.7 m/s. It strikes the first block when the latter is at the rightmost point in its oscillation. The collision is completely inelastic, and the two blocks stick together.

a) Determine the frequency of the resulting motion.
I have found this to be 0.58 Hz, and this is the answer.

b) Determine the amplitude of the resulting motion.
E = (1/2)kx^2 + (1/2)mv^2 = 1/2k(A_new)^2 (I think?)
kx^2 + mv^2 = k(A_new)^2
(27)(.1)^2 + (.8)(1.7)^2 = (27)(A_new)^2
A_new = .332

This is what I get, but it does not seem right, and is not accepted as a valid answer... If my approach is wrong, then how do I solve this part?

c) Determine the phase constant (relative to the original t = 0) of the resulting motion.

Well, the relevant equations, I suppose, are:

x = A cos(wt + phi)
v = -w A sin (wt + phi)
a = -w A cos (wt + phi)

However, even once I have the amplitude, I don't know how to solve this part of the question. Please give a hint or something. Thanks.
 

Answers and Replies

  • #2
alphysicist
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Hi akan,

Homework Statement


A 1.2-kg block rests on a frictionless surface and is attached to a horizontal spring of constant k = 27 N/m. The block is oscillating with amplitude 10 cm and phase constant phi = -pi/2. A block of mass 0.80 kg is moving from the right at 1.7 m/s. It strikes the first block when the latter is at the rightmost point in its oscillation. The collision is completely inelastic, and the two blocks stick together.

a) Determine the frequency of the resulting motion.
I have found this to be 0.58 Hz, and this is the answer.

b) Determine the amplitude of the resulting motion.
E = (1/2)kx^2 + (1/2)mv^2 = 1/2k(A_new)^2 (I think?)
kx^2 + mv^2 = k(A_new)^2
(27)(.1)^2 + (.8)(1.7)^2 = (27)(A_new)^2
A_new = .332
When they refer to the resulting motion, they mean the motion after the collision. Immediately after the collision, what is the total mass on the spring, and what speed does it have? (You have 0.8kg and 1.7m/s, but that is the mass of the moving block and its speed just before the collision.) Once you correct that, I believe you will get the right answer.
 
  • #3
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m1v1 + m2v1 = v2(m1+m2).
m2v1 = v2(m1+m2)
v2 = m2v1 / (m1+m2)

And this is the velocity I use in the equation, right? Is everything else correct? Thanks.
 
  • #4
alphysicist
Homework Helper
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m1v1 + m2v1 = v2(m1+m2).
m2v1 = v2(m1+m2)
v2 = m2v1 / (m1+m2)

And this is the velocity I use in the equation, right? Is everything else correct? Thanks.
That looks like the right idea. Use that velocity and the total mass.
 
  • #5
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That gave me the correct result. However, how do I solve part C? I guess what I don't undersand is what the wt term, and what do I substitute in there. I understand that t is a point in time, but what's it equal to? w is an angular velocity, but what is that equal to? The angular velocity of the system or at the point of time? Thanks.
 
Last edited:
  • #6
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Ok, so x = A cos (wt + phi)
If we take the new t = 0 as the moment of collision, phi = arccos(x/A) = arccos((.10)/.21) = 58.41186449.
But relative to the original t = 0, which had a constant of -pi/2 = -90 degrees, we have it that the new phi = 58.41186449 - 90 = -28.44. But that's not accepted as a valid answer. Any hints?
 
Last edited:

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