Naked Science Forum
General Science => General Science => Topic started by: rmolnav on 13/09/2017 10:54:34

Following question was sent in 2010, which I´ve seen a few days ago.
"Why (is) a pendulum is affected by gravity and a mass on a spring is not?"
That question was basically wrong, because the oscillatory movement of an object hanging from a spring is the result both of gravity (constant weight always acting downwards), and the oscillatory upward tension of the spring ...
Unless spring own weight were insignificant (relative to the mass of the object), that weight should also be taken into consideration to get an accurate movement equation.
A reasonable question should rather be the one of current "subject".
Any comments?

Are you saying answers to original question were wrong?
I would have expected ansers to the original question to address the point you raise.

Thank you.
I say that the very question was erroneous. And there were just a few answers, most not realizing that fact ... :
https://www.thenakedscientists.com/forum/index.php?topic=35444.0

I believe that the massonaspring problem is usually set up in the horizontal plane, so there is no change in gravitational potential energy associated with spring oscillation. (in the simplest case, this assumes a perfectly rigid spring orthogonal to the axis of the spring or a frictionless surface along which the mass slides)
For a vertical system gravity would seem to be an essential part of the problem.

Unless spring own weight were insignificant (relative to the mass of the object), that weight should also be taken into consideration to get an accurate movement equation.
A reasonable question should rather be the one of current "subject".
Any comments?
Although that is true and the weight of the spring would affect the period, I think the original question is valid. The OP is asking why the period of oscillation of a spring/mass system is unaffected by gravity whereas a pendulum period changes with gravity.
I don't think the question was answered properly

For a large pendulum, perhaps 10m long, the force of gravity does not change significantly over the height change. So the mass is subjected to an approximately constant force towards the center of the Earth. The motion is sinusoidal.*
For a large spring, perhaps 10m long, the spring will extend until the weight of the spring matches the force from the extension of the spring. If you now set this mass vibrating vertically, with the same amplitude as the pendulum, the spring force will vary significantly as the spring gets longer and shorter. So this vertical motion will be very different from a sine wave.
* Note that motion of a pendulum is only sinusoidal for small angles. Beyond this, some nonsinusoidal terms become significant.

For a large spring, perhaps 10m long, the spring will extend until the weight of the spring matches the force from the extension of the spring. If you now set this mass vibrating vertically, with the same amplitude as the pendulum, the spring force will vary significantly as the spring gets longer and shorter. So this vertical motion will be very different from a sine wave.
While that's true I don't think it is what the original OP https://www.thenakedscientists.com/forum/index.php?topic=35444.0 (https://www.thenakedscientists.com/forum/index.php?topic=35444.0) was asking.
Most explanations of spring/mass assume perfect spring or small oscillation so there is no variation of k with extension.
For a pendulum the restoring force is mg (vector) so if you keep the length and mass constant but change g by taking the pendulum to the top of a mountain then the period will change. However, for a spring the restoring force is spring constant x displacement from equilibrium so g doesn't play a part and for a perfect spring/small oscillation the period is the same whether horizontal, vertical or for all values of g. That I think is what was puzzling the original OP.

#3
In that case of movement somehow obliged to be horizontal, it´s clear what you say. Or, considering forces, as gravity would be perpendicular to the movement , it would not produce any acceleration of the mass ...
But real surfaces are not completely frictionless. There would be some negative acceleration, proportional to mass weight, small effect of gravity.
In any case, the comparison and question would be rather absurd, as far as I can ee.

#4
If original question meant "not affected by g acceleration changes", instead of just " not affected by gravity", perhaps it was not answered properly because it was not understood that way.
In that case, I would say that pendulum tangential acceleration, for a given weight, changes only due to a geometrical reason: tangential component of the oscillating mass weight changes sinusoidally. As weight changes with g, so do tangential accelerations
But in the mass/vertical spring case, two opposite forces affect the movement of the mass: downwards, its constant weight; and upwards, the tension of the spring, proportional to its relative stretch. And if g acceleration changes, BOTH forces change in same proportion, and actual mass acceleration changes same way as in first case with different g ...

If original question meant "not affected by g acceleration changes", instead of just " not affected by gravity", perhaps it was not answered properly because it was not understood that way.
I suspect the original poster was not familiar enough with the subtleties and should have said "not affected by changes in the value of g"
But in the mass/vertical spring case, two opposite forces affect the movement of the mass: downwards, its constant weight; and upwards, the tension of the spring, proportional to its relative stretch.
Yes, with a change of g the mass finds a new equilibrium point and the acceleration of the mass will be dependent only on the deflection and spring constant, not the value of g, hence the period does not change with changes in value of g.
As I said, I think this is the real question that is being asked in https://www.thenakedscientists.com/forum/index.php?topic=35444.0 (https://www.thenakedscientists.com/forum/index.php?topic=35444.0) as the OP mentions time period in the opening post.

But real surfaces are not completely frictionless.
True, but "real" springs have mass throughout, and have variable spring "constants" ("real" springs are anharmonic oscillators). And any "real" pendulum or spring calculation for a system operating on the Earth must also take the Coriolis Effect into account. Also, there are some relativistic terms we forgot to include, and oh yeah the spring is likely to be ferromagnetic and interacting with Earth's magnetic field...
See where I'm going with this?
It is very common for such physics problems to ignore friction along with countless other terms. (hence the joke about the physicist invoking a "spherical cow in a vacuum" when solving a problem on agronomy). One must select the degree of precision required in the solution, and chose a model with enough complexity to achieve that precision, but so much as to make the calculation unnecessarily complicated.

#5
I consider very exaggerated to speak about possible changes in gravity due to height change, if the pendulum is not long, because if f.e. of 5 or 10 m. the mass wouldn´t be "subjected to an approximately constant force towards the center of the Earth” …
As said in #10 "One must select the degree of precision required in the solution", and I´m afraid mentioned detail goes beyond practically any case required precision ...

#10
I agree with you.
Regarding terms mentioned in "to ignore friction along with countless other terms", we could bring up site latitude, Moon´s position at the moment, GMT, altitude, near or far from from big masses (mountains, amounts of ice there ...), actual composition of Earth´s crust in the area ...
Nature is far more complex than our simplifications !!!

For a pendulum the restoring force is mg
The restoring force on a pendulum is mgsinϴ.
Since sinϴ ≈ ϴ for small ϴ, then F ≈ mgϴ, that is, the restoring force is proportional to the displacement, which is the condition for simple harmonic motion.
The displacement of a spring is also proportional to the displacement (Hooke's Law), so that also produces SHM. The effect of gravity on the spring/mass is not proportional to the displacement, so it doesn't produce SHM.

For a pendulum the restoring force is mg
The restoring force on a pendulum is mgsinϴ.
I think you are slightly misquoting me, what I said was "For a pendulum the restoring force is mg (vector)", and yes that vector component is mgsinϴ.
Since sinϴ ≈ ϴ for small ϴ, then F ≈ mgϴ,
But that isn't always true.
You are making an approximation. As @chiralSPO said, we often make approximations in order to simplify calculations, and that is valid in both practical and teaching situations.
For a pendulum Sinϴ ≈ ϴ is only useful for ϴ<10°, although at 20° the error is only around 1%, for ϴ>22° the difference is noticeable.
Similarly for the spring/weight I did say "for a perfect spring/small oscillation" and so using a similar small deflection the variation of g is considered to be zero.
If you use approximations for one you ought to allow them for the other.

#13 & #14
If for a pendulum we simplify considering Sinϴ ≈ ϴ (small angles), the motion can be considered as simple harmonic, that´s clear ...
But please note I have not mentioned that condition before, because I just had said (#4) that "tangential component of the oscillating mass weight changes sinusoidally".
I have a curious animated archive, which I´m not sure it could be included and seen here. i´ll try:
Oscillating_pendulum.gif

#15 (continuation)
It does not work directly, but googling the "pseudolink" it appears, along with many others ... Now is placed page first place ...

For a large pendulum, perhaps 10m long, the force of gravity does not change significantly over the height change. So the mass is subjected to an approximately constant force towards the center of the Earth. The motion is sinusoidal.*
For a large spring, perhaps 10m long, the spring will extend until the weight of the spring matches the force from the extension of the spring. If you now set this mass vibrating vertically, with the same amplitude as the pendulum, the spring force will vary significantly as the spring gets longer and shorter. So this vertical motion will be very different from a sine wave.
* Note that motion of a pendulum is only sinusoidal for small angles. Beyond this, some nonsinusoidal terms become significant.
Are you sure about that?
The requirement for shm is that the restoring force is proportional to the displacement.
For any normal spring that's going to be true. That's Hoke's law and most things obey it quite well. Springs are specially designed to follow it particularly closely.
For a pendulum it's only strictly true in the case of zero displacement.
I suspect that for the sort of thing you would set up in a school lab, the spring would be closer to a sine wave than the pendulum.

I think you are slightly misquoting me, what I said was "For a pendulum the restoring force is mg (vector)"
I thought you were saying "For a pendulum the restoring force is mg (mg is a vector)....."
Since sinϴ ≈ ϴ for small ϴ, then F ≈ mgϴ,
But that isn't always true.
If it's left open for the user to decide how approximately equal, and how small, it can't not be true.You are making an approximation. As @chiralSPO said, we often make approximations in order to simplify calculations, and that is valid in both practical and teaching situations.
For a pendulum Sinϴ ≈ ϴ is only useful for ϴ<10°, although at 20° the error is only around 1%, for ϴ>22° the difference is noticeable.
Similarly for the spring/weight I did say "for a perfect spring/small oscillation" and so using a similar small deflection the variation of g is considered to be zero.
If you use approximations for one you ought to allow them for the other.
The question was why doesn't gravity affect the springmass system, and the answer is because gravity exerts a constant force and not one dependent on displacement. That's true regardless of any approximation in the linearity of the pendulum restoring force, and/or Hooke's law.

#18
"The question was why doesn't gravity affect the springmass system, and the answer is because gravity exerts a constant force and not one dependent on displacement. That's true regardless of any approximation in the linearity of the pendulum restoring force, and/or Hooke's law."
[/quote]
You apparently mean gravity could affect the springmass system only if the force exerted by it changed with time in a given case ... But on another spot of our planet surface, a different g would affect the system IF Hooke´s law were different ...
As said on #8, the fact that upward restoring force would change at same proportion as weight IS the reason why motion main details would not change even changing location.

The question was why doesn't gravity affect the springmass system, and the answer is because gravity exerts a constant force and not one dependent on displacement.
Sorry, when I read your post, I thought you were implying that because gravity is not constant with height it does affect the spring/mass system  "The effect of gravity on the spring/mass is not proportional to the displacement, so it doesn't produce SHM"  and hence the spring/mass doesn't produce SHM. I think we are talking at cross purposes here because your comment "the answer is because gravity exerts a constant force and not one dependent on displacement" means we are in agreement because the restoring force on the mass therefore only varies with x, assuming perfect spring.