# Is it SHM or not?

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1. Jan 19, 2016

### SANGEETAMILIND

In an SHM, the only force that should be acting, that is the net force should be the restoring force F, by definition...
F = -kx

For example there is a massless spring of spring constant k attached to the ceiling and there is a body of mass m hung at it and avoiding all kinds of friction...

Due to the weight of the body the equilibrium mean position will be shifted...
numerically it would become
xo = (mg/k)
{considering the the original mean position to be 0}

angular frequency and time period would still be the same as in the SHM motion performed by the spring if it were kept horizontally at natural length.

But by definition of Simple Harmonic Oscillators
F= -kx should be the only force... and in the vertically hung spring mass system 'mg' is also acting, not withstanding that the motion is exactly like the SHM with a different equilibrium position.
Would it still be considered as a simple harmonic oscillator?

Last edited: Jan 19, 2016
2. Jan 19, 2016

### mfig

Your definition of SHM is not quite right. Simple harmonic motion is a variety of periodic motion that results when the restoring force is proportional to displacement from equilibrium. The gravitational force is not a restoring force.

3. Jan 19, 2016

### jbriggs444

Yes, this still gives simple harmonic motion. The net force is still given by F = -kx where x is the deviation from the equilibrium position. The fact that the net force is a combination of the spring force plus gravity and the fact that the equilibrium length of the spring is not the same as its unstressed length is irrelevant.

4. Jan 19, 2016

### SANGEETAMILIND

For a Simple Harmonic Oscillator...

the net force is the restoring force which is F ∝ -x.

But in the above scenario, the net restoring force is the combination of Gravity and Spring Force and thus not directly proportional to displacement...

So would it be called a Simple Harmonic Oscillator or just a Harmonic Oscillator?

5. Jan 19, 2016

### Staff: Mentor

True.
How do you figure that, mathematically?

6. Jan 19, 2016

### SANGEETAMILIND

the net force is
F = -(kx + mg)

can you call it as F ∝ -x?

7. Jan 19, 2016

### jbriggs444

Yes, it is directly proportional to displacement from the equilibrium position.

8. Jan 20, 2016

### Staff: Mentor

OK, let's go through this step by step. We have a mass hanging vertically from a spring under the influence of gravity. If we could "turn off" gravity, the mass would "hang" at a certain vertical position corresponding to the equilibrium length of the spring alone. Call this vertical position x = 0. Let x be positive upwards and negative downwards.

Now "turn on" gravity. It pulls the mass down. Let it come to rest at a new equilibrium position, call it x = x0. At this position the net force on the mass is zero: Fnet = -kx0 - mg = 0. Note that x0 is negative so -kx0 is positive (upwards). The spring pulls the mass upwards and gravity pulls the mass downwards.

Now pull the mass down to a new position x and release it. Immediately after release:

(a) What is Fnet on the mass now?

(b) What is the displacement from the equilibrium position (the one with gravity turned on)?

(c) Write Fnet in part (a) in terms of the displacement in part (b).

9. Jan 20, 2016

OK.