# Simple Harmonic Motion vibrations

## Homework Statement

Find the period of low-amplitude vertical vibrations of the system shown. The mass of the block is m. The pulley hangs from the ceiling on a spring with a force constant k. The block hangs from an ideal string.
http://www.luiseduardo.com.br/undulating/SHM/shmproblems_arquivos/image105.jpg [Broken]

Fel = K.x
Fr = m.a

## The Attempt at a Solution

Hello,

This kind of problem is difficult for me, because I don't understand very well the "geometry" of the problem. When the block falls a distance "x", what is the compression of the spring? Please, could anyone help me ?
Thanks.

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Chestermiller
Mentor

## Homework Statement

Find the period of low-amplitude vertical vibrations of the system shown. The mass of the block is m. The pulley hangs from the ceiling on a spring with a force constant k. The block hangs from an ideal string.
http://www.luiseduardo.com.br/undulating/SHM/shmproblems_arquivos/image105.jpg [Broken]

Fel = K.x
Fr = m.a

## The Attempt at a Solution

Hello,

This kind of problem is difficult for me, because I don't understand very well the "geometry" of the problem. When the block falls a distance "x", what is the compression of the spring? Please, could anyone help me ?
Thanks.
You really need to keep track of 2 x's in this problem: x1 = distance of center of pulley from top, x2 = distance of mass from top. You need to be able to express the change in the length of each of the springs in terms of the changes in these two distances. This is a kinematics (geometry) problem that must be solved first before considering the mechanics.

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Well this one is a bit confusing, I agree. What happens to the rest of the system, if say, you compress the upper most spring?

Chestermiller
Mentor
Let's talk about the kinematics. If you move the pulley down by Δx1 relative to the equilibrium position of the system (and you prevent the pulley from rotating), the upper spring stretches by Δx1, the mass moves downward by Δx1, and the lower spring compresses by Δx1. If you move the mass down by Δx3 relative to its location in the equilibrium position with respect to the pulley (say, by preventing the pulley from translating), the lower spring stretches Δx3. So

Stretch of upper spring = Δx1

Downward movement of mass = Δx1+Δx3

Stretch of lower pulley = Δx3 - Δx1

Now let Δx2 = Δx1+Δx3= total amount that mass moves down relative to equilibrium position

Then Δx3 = Δx2-Δx1

Stretch of upper spring = Δx1

Downward movement of mass = Δx2

Stretch of lower pulley = Δx2 - 2Δx1

When the spring/mass system experiences these displacements relative to the equilibrium position of the system,
(a) What is the tension in the upper spring?
(b) What is the tension in the lower spring (and in the spring)?
(c) Taking the pulley as a free body, how is the tension in the upper spring related to the tension in the lower spring and string?
(d) Express Δx1 in terms of Δx2.
(e) What is the tension in the lower spring in terms of Δx2 alone?
(f) What is the tension in the spring?
(g) From a free body diagram on the mass, obtain a newton's 2nd law force balance on the mass involving only Δx2

1 person
Well, chestermiller, I tried what you suggested, please, correct me:

a) Fel1 = K.Δx1
b) Fel2 = K.(Δx2-Δx1)
c) Fel1 = Fel2 + T
d) Δx1 = Δx2/3 + T/3K
e) Fel2 = K(Δx2/3 - 2T/3K)
f) mg - T -Fel2 = m.a

I'm a bit confused :|

Chestermiller
Mentor
Well, chestermiller, I tried what you suggested, please, correct me:
Good try. You have the correct general idea.
a) Fel1 = K.Δx1
b) Fel2 = K.(Δx2-Δx1)
Previously we said that the change in length of the lower spring relative to the system equilibrium point is (Δx2-2Δx1), so
Fel2 = K.(Δx2-2Δx1)
where Fel1 and Fel2 are the changes in the spring tensions relative to the system equilibrium point.
c) Fel1 = Fel2 + T
The string and lower spring are connected to one another over a massless frictionless pulley. So, the tension change in the string T (relative to the system equilibrium point) has to be the same as the tension change in the lower spring (relative to the system equilibrium point). So T = K.(Δx2-2Δx1), and
K.Δx1=2K.(Δx2-2Δx1)
So, 5Δx1=2Δx2
or Δx1=0.4Δx2
If we substitute this back into our equation for T, we get:
T = 0.2KΔx2

Relative to the system equilibrium point, the force balance on the mass is:

ma = -T = -0.2KΔx2

The mg term should not be in this equation because we are looking at the change in the system relative to the equilibrium point. The acceleration of the mass is :
$$a=\frac{d^2(Δx_2)}{dt^2}=-\frac{0.2K}{m}(Δx_2)$$

Given that this is the differential equation for the motion of the mass, what is the frequency of the oscillations? What is the period?

You definitely had the right idea when you approached this problem, but, it's a pretty hard problem. You had to be able to dope out the kinematics of the motions, and even realize that there are two displacement locations that you need to consider. You also had to realize that you were dealing with incremental changes, rather than absolute locations. You also had to realize that the tension of the string is the same as the tension in the lower spring. You did a good job of identifying the free bodies and carrying out the force balances. I regard doping out the kinematics as the most difficult part of this problem

Chet

1 person
Thank you very much Chestermiller !!!

Amlitude application of car

When amplitude of SHM is constant(assuming no friction) why this amplitude is analogous to car springs displacement? Because after force applied car spring stretches (in real applications) and then displacement continues to decrease over a period of time. I am confused. Can anyone explain please.

Chestermiller
Mentor
When amplitude of SHM is constant(assuming no friction) why this amplitude is analogous to car springs displacement? Because after force applied car spring stretches (in real applications) and then displacement continues to decrease over a period of time. I am confused. Can anyone explain please.
In a car, you have shock absorbers to provide damping, so the amplitude decreases. If you took the shock absorbers out, the oscillation would continue for a much longer time.

1 person
In a car, you have shock absorbers to provide damping, so the amplitude decreases. If you took the shock absorbers out, the oscillation would continue for a much longer time.

Thank you so much...