Why a spring-mass assembly is a SHM?

[jaumzaum]

I was trying to understand why a spring -mass assembly and a simple pendulum is a simple harmonic movement.

I've done like this:

In a SHM that starts in -A, we have x = -A.cos(wt)

In a spring-mass we have that the maximum velocity is $\frac{mv²} {2} = \frac{kA²}{2} -> v = A\sqrt{\frac{k}{ m}}$

$w = v/A = \sqrt{\frac{k}{m}}$

$v = \int a.dt = \int \frac{-k.x}{m} dt$

But in a SHM, $v = -w .x.tan(wt) -> -w .x.tan(wt) = \int \frac{-k.x}{m} dt$


If we substitute x = -A.cos(wt), we see that the expression is true, so spring-maass is a SHM.

But I don't know how to demonstrate why pendulum is a SHM?

 

[rock.freak667]

Well at an angle θ, what the torque about the center of rotation equal to? (use m,g, and L for parameters)

This torque is also equal to I(d²θ/dt²). 'I' in this case just the inertia of the mass.

 

[PeterO]

I was trying to understand why a spring -mass assembly and a simple pendulum is a simple harmonic movement.

I've done like this:

In a SHM that starts in -A, we have x = -A.cos(wt)

In a spring-mass we have that the maximum velocity is $\frac{mv²} {2} = \frac{kA²}{2} -> v = A\sqrt{\frac{k}{ m}}$

$w = v/A = \sqrt{\frac{k}{m}}$

$v = \int a.dt = \int \frac{-k.x}{m} dt$

But in a SHM, $v = -w .x.tan(wt) -> -w .x.tan(wt) = \int \frac{-k.x}{m} dt$


If we substitute x = -A.cos(wt), we see that the expression is true, so spring-maass is a SHM.

But I don't know how to demonstrate why pendulum is a SHM?

The condition for SHM is that the restoring force is proportional to displacement.
That is pretty easy to do with a spring/mass set-up.

For a pendulum, when the bob is off to one side, the component of the mass, in the direction of travel, is the restoring force. This involves a trig function, but we can show that for small deflections, the restoring force is pretty much proportional to displacement.

 

[jaumzaum]

Thanks,

If we calculte we get F = mg sin(y)cos(y)

where sin(y) = x/L if cos(y) ~ 1, we get a SHM




So in big angles, pendulum is not a SHM?

 

[PeterO]

Thanks,

If we calculte we get F = mg sin(y)cos(y)

where sin(y) = x/L if cos(y) ~ 1, we get a SHM

So in big angles, pendulum is not a SHM?

Not sure where that cos function came from?? but yes for small angles

Tan(x) = x = sin(x) [x in radians of course]

So this is SHM for "small" angles. The conjecture then becomes "what is small?" And that can make the subject of a great little practical exercise.

 

[jaumzaum]

mg sin(y) would be the force tangent to the movement, so the force in the horizontal (x) axes is mgsin(y)cos( y)