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Slowing a Mass with a Spring Collision

  • Thread starter Coop
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Homework Statement



A 1.5 kg mass slides across a horizontal, frictionless surface at a speed of 2.0 m/s until it collides with and sticks to the free end of a spring with spring constant 50 N/m. The spring's other end is anchored to the wall. How far has the spring compressed when the mass, at least for an instant, is at rest? How much time does it take for the spring to compress to this point?

I don't care about the first part of the question. I am just looking at the second question.

Homework Equations



[tex]T (period) = 2\pi\sqrt{\frac{m}{k}}[/tex]

m = mass on the spring
k = spring constant

The Attempt at a Solution



The answer in my book says to set [tex]t_{f} = \frac{1}{4}T[/tex], stating "the motion of the mass until it stops is 1/4 of a cycle of simple harmonic motion." This is why I am confused, isn't the spring only momentarily at rest at the beginning, end and halfway through the period? Isn't the book saying the spring is momentarily at rest one-forth of the way through the period? I don't see why this is true.

Thanks,
Coop
 

Answers and Replies

  • #2
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In this particular problem, the mass is moving at its top speed when it first contacts the spring and sticks to it. In SHM, the top speed is attained at the equilibrium position. So the point where the mass sticks to the spring will eventually become the equilibrium position. 1/4 of a cycle later, it has slowed to zero, and the spring begins to rebound. 1/4 cycle later, it is again passing through the equilibrium position, and exhibiting the maximum velocity in the other direction; also, at this point, the spring is no longer compressed. During the next 1/4 cycle, the spring begins stretching out, and the mass starts to slow down again, with the mass moving in the direction opposite to its original direction. By the end of this 1/4 cycle, the velocity is again zero, and the string exhibits its maximum stretch. During the next 1/4 cycle, the mass moves in its original direction again, speeding up. By the end of this 1/4 cycle, it again reaches the equilibrium position with maximum velocity.
 
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In this particular problem, the mass is moving at its top speed when it first contacts the spring and sticks to it. In SHM, the top speed is attained at the equilibrium position. So the point where the mass sticks to the spring will eventually become the equilibrium position. 1/4 of a cycle later, it has slowed to zero, and the spring begins to rebound. 1/4 cycle later, it is again passing through the equilibrium position, and exhibiting the maximum velocity in the other direction; also, at this point, the spring is no longer compressed. During the next 1/4 cycle, the spring begins stretching out, and the mass starts to slow down again, with the mass moving in the direction opposite to its original direction. By the end of this 1/4 cycle, the velocity is again zero, and the string exhibits its maximum stretch. During the next 1/4 cycle, the mass moves in its original direction again, speeding up. By the end of this 1/4 cycle, it again reaches the equilibrium position with maximum velocity.
Oh I see, I was thinking the spring was starting from it's maximum amplitude, but it's not; it's starting at its eq. point. Thanks :)
 

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