Why Does Spring Oscillate Horizontally?

  • Thread starter Thread starter Atomos
  • Start date Start date
  • Tags Tags
    Resonance Spring
Click For Summary
An oscillating spring tends to shift to horizontal oscillation due to its natural tendency to twist when stretched, creating a sideways force that enhances side-to-side motion. Increased mass reduces this twisting effect because heavier masses have greater inertia, making them less responsive to the spring's twisting. The interaction between torsional oscillation and vertical oscillation can lead to complex behaviors, such as alternating modes of oscillation. Additionally, the flexing of support rods contributes to horizontal forces, but excessive weight limits oscillation amplitude and frequency. Understanding the relationship between mass, frequency, and the resulting forces is crucial in explaining these dynamics.
Atomos
Messages
165
Reaction score
0
I am asked to explain why an oscillation spring will eventually begin oscillating in the horizontal direction, i.e. become a pendulum, and I am asked why a mass that is too great will not become a pendulum.

I know the answer to the first part of the question. The spring has a natural tendency to twist when stretched. This twisting produces a small force in the sideways direction that adds to the amplitude of the side to side oscillation every time the spring is fully stretched. However, I am not sure why this effect is reduced or eliminated with increased mass. Is it because the heavier mass has more inertia and thus does not twist as easily?
 
Physics news on Phys.org
Careful ... twisting of the spring produces torque around a vertical axis,
but this is not really an external horizontal Force. It would result in a
significant "Torsional oscillation" if the torque / rotational inertia has
the same natural frequency of oscillation as the vertical oscillation mode. With care, these modes can be made to alternate (google "wilberforce").
If the mass is too light, vertical motion is too frequent to match the twist
but if the mass is too heavy the vertical motion is not frequent enough.

External horizontal Force (if that's what you want) is usually mostly
the flexing of the support rods. With too heavy a weight, the amplitude
of oscillation will be low (to NOT hit the floor); a slow frequency will make small variation in acceleration, so won't change the post flexure much
(what causes Work to be done by the post: not just F, but F dx !).
hope this helped
 
Last edited:
I don't really understand how a slower frequency of the application of the sidways motion (i.e. bending of support) would prevent the pendulum motion. Even if it is slow, at certain masses, it will occur at such multiples of time that is adds to the amplitude of the side to side motion.
 
Last edited:
And why would a greater mass cause the amplitude of the bending support to decrease?
 
bumpedy bump bump
 
you have z(t) = A sin(wt) , with height variation 2A.

Now compute the acceleration function ...
what's the variation in the acceleration?
So what's the variation in the Force applied to the support?
Does it depend on angular frequency? does it depend on m?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
5
Views
988
Replies
6
Views
960
Replies
17
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
1K
Replies
17
Views
3K
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
731
  • · Replies 20 ·
Replies
20
Views
4K