Conceptual Spring Problem Help requested

  • Thread starter Rmehtany
  • Start date
  • #1
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Homework Statement


A mass hangs from the ceiling of a box by an ideal spring. With the box held fixed, the mass is given an initial
velocity and oscillates with purely vertical motion. When the mass reaches the lowest point of its motion, the box
is released and allowed to fall. To an observer inside the box, which of the following quantities does not change
when the box is released? Ignore air resistance.

(A) The amplitude of the oscillation
(B) The period of the oscillation <- CORRECT
(C) The maximum speed reached by the mass
(D) The height at which the mass reaches its maximum speed
(E) The maximum height reached by the mass

No variables; this is a conceptual problem

Homework Equations


A_k = -kx/m
A_g = mg
Period = 1/2*pi * sqrt(m/k)

The Attempt at a Solution


I tried to use the two accelerations and set them equal and see what happens:

-kx/m = mg
g = -kx/m^2

I am confused, I have never seen such a problem, and I don't know what to do. Can someone lead me in the right direction?
 

Answers and Replies

  • #2
OldEngr63
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You need to do two things:
1. Write the differential equation of motion for the mass in the falling box;
2. Determine the initial conditions for the DE at the instant that the box starts to fall.

The first step will be facilitated by drawing a FBD.
 
  • #3
27
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What if i don't know calc

I drew a FBD with two forces
 
  • #4
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With no calculus knowledge, you are probably in over your head.
 
  • #5
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The question was on my AP Physics B hw, and we haven't learned calculus. I wanted to know how to do it
 
  • #6
haruspex
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You can resolve some of them, at least, without calculus.
The formula you have for the period does not involve g. It follows that the period would be the same with a different value for g. In free fall, g is effectively 0, and nothing else has changed that affects the period.
For amplitude, you know that the equilibrium extension of the spring is shorter (0 instead of mg/k). If the amplitude were the same, what would the two maximal extensions be? What is the actual extension just after the string is cut?
 
  • #7
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Oh, I see, so the only thing that doesn't change is period because it doesn't depend upon g

Amplitude changes from mg/k to how far the object falls. The object potentially could reach a new max speed assuming the height the observer dropped is far enough. If the max speed changes, the height at which it occurs changes. But wouldn't the max height remain the same?
 
  • #8
haruspex
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Amplitude changes from mg/k to how far the object falls
You don't know what the amplitude was before the string was cut. An extension of mg/k would have been the equilibrium position. If the amplitude initially was A, what would the greatest and least extensions have been?
 
  • #9
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Would the greatest extension be height fallen + A + equilibrium position height? And the least be equilibrium position height - A?
 
  • #10
haruspex
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Would the greatest extension be height fallen + A + equilibrium position height? And the least be equilibrium position height - A?
I don't think that's quite what you meant. The greatest extension is not (whatever) + a height. Put it in terms of greatest extension as (whatever) + equilibrium extension. Then substitute in the known value of equilibrium extension.
When you've got that, try to answer these questions:
- what is the extension at the moment the string is cut?
- what is the new equilibrium extension?
- what does that tell you for the new amplitude?
 

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