# Why don't we consider mg?

1. Oct 3, 2012

Dear All!
I have attached the question. Please explain as to why we are ignoring mg when we are taking equation of motion for the spring moving in the vertical axis?
Thank you

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2. Oct 3, 2012

### Staff: Mentor

Welcome to the PF.

It must be part of the original problem statement. You are correct that if that system is oriented vertically in a gravitational field, there will be a force term due to gravity as part of the equation. Perhaps the problem stipulates that the system is oriented horizontally on a frictionless surface?

3. Oct 3, 2012

### jbriggs444

If you add a fixed vertical force such as gravity to the force equation, you still end up with simple harmonic motion. The centerline of the motion ends up displaced downward a tad and some of the approximations being used get pushed toward the edge of their validity.

So gravity just complicates the example without adding anything useful. Pretend the setup is horizontal on an air table if you like.

4. Oct 3, 2012

### AlephZero

Probably because you have to learn any subject one step at a time. Adding the weight into the differential equation would make things more complicated, and obscure the point of the exercise.

In real life, if the springs were actually a tightly stretched wire, the weight of the object would be negligible compared with the other forces acting on it. For example if you are studying the vibration frequency of a guitar string, you can ignore whether the orientation of the guitar is horizontal or vertical.

5. Oct 3, 2012

### HallsofIvy

Staff Emeritus
If you have an object of mass m, a spring of natural length l and spring constant k, hanging vertically, then the equation of motion is
$$m\frac{d^2x}{dt^2}= -k(x- l)- mg$$
which is the same as
$$m\frac{d^2x}{dt^2}= -kx- (kl+ mg)$$

Notice that, without any motion at all, the gravitational pull on the mass would extend the length of the spring until the two forces offset: mg= k(x- l) or x= (mg+ kl)/k. If we took that to be the natural length of the spring, and ignored gravity, we would have
$$m\frac{d^2x}{dt^2}= -k(x- (mg+kl)/k)= -kx- (kl+ mg)$$

In other words, adding gravity to such a problem is exactly the same as changing the 'natural length' of the spring from what we would see if the spring were lying on a horizontal surface to what it is hanging.

6. Oct 3, 2012

### AlephZero

The situation here is different from what Halls described. The force causing the oscillation does not come from the change in tension in the springs, but from the change in direction of the tension force.

To linearize the problem we asssume the tension is constant, which is the same as ignoring change in length from $l$ to $\sqrt{l^2 + y^2}$ of the springs.

If the weight gives an equilibrium position at a non-zero angle $\bar\theta$, the period for small oscillaations about that position is different by a factor involving $\cos\bar\theta$.

But it is quite realistic for the tension T to be several orders of magnitude greater than the weight mg, so $\bar\theta$ is so small it (and the weight) can be ignored.

In "real world" vibration situations it is quite common for objects to have accelerations much bigger than 1g. For example a guitar string vibrating at 1000 rad/s (about 160 Hz) with an amplitude of 2mm peak to peak in the centre has an acceleration of about 100g. For practical purposes, the effect of an extra "1g" from the weight of the object is often negligible.

7. Oct 4, 2012

### sophiecentaur

A mass on a single uniform spring will perform simple harmonic motion for large amplitudes but, in the arrangement in the OP (and also the simple pendulum), the restoring force is not actually proportional to displacement and the SHM approximation is only for small angles (a single sine wave). For the 'two springs' in the absence of gravity, the displacement/ time graph will be symmetrical so the harmonic content of the curve will be only even order harmonics but, when you include the effect of gravity, there will also be odd order harmonics in the curve.

8. Oct 4, 2012

Thank you so much everyone! It makes sense now =)

9. Oct 4, 2012

Could you please explain what you mean by the above line?

10. Oct 4, 2012