Derivation: Simple Harmonic Motion

[serverxeon]

I know how to arrive at

a = -(k/m)x

for a SHM system.

However, most derivations i found did an unexplained step from here onward.

They simply let k/m = ω².

And later on then mention ω as the angular frequency and link to the period of oscillation.

My question is,

Why did that substitution take place? Why must it be ω²?
Why can't k/m = v³ or anything else?

Please help as I couldn't find any explanation for this step!

 

[maxwellcauchy]

a = d²y/dt² = v*(dv/dx)
Also you have kx + ma = 0, then substitute a with v*(dv/dx), then solve the first ODE.

 

[serverxeon]

hi, thanks for the respond.

could u please explain how you arrive at

a = d2y/dt2 = v*(dv/dx)?

why is there 'y' and 'x' and why dv/dx is multiplied by v?

 

[technician]

A physics definition of SHM is the motion that results when an object experiences a restoring force when it is displaced.
SHM is when the restoring force is proportional to the displacement.
i.e F proportional to x, F= kx (k is a constant)
This means that k = F/x which has units N/m and is known as the STIFFNESS (same term is used for springs)
If the object has mass, m then acceleration = F/m = (kx)/m
By making comparisons with circular motion where a = ω^2 A (A is radius which becomes amplitude in oscillations) it is possible to show that ω^2 = k/m
The period of the circular motion is the same as the period of the SHM
This means that ω = √k/m or 2πf = √k/m
so you see that straight away you can calculate the frequency of the SHM from the stiffness and mass of the system.

 

[serverxeon]

A physics definition of SHM is the motion that results when an object experiences a restoring force when it is displaced.
SHM is when the restoring force is proportional to the displacement.
i.e F proportional to x, F= kx (k is a constant)
This means that k = F/x which has units N/m and is known as the STIFFNESS (same term is used for springs)
If the object has mass, m then acceleration = F/m = (kx)/m
By making comparisons with circular motion where a = ω^2 A (A is radius which becomes amplitude in oscillations) it is possible to show that ω^2 = k/m
The period of the circular motion is the same as the period of the SHM
This means that ω = √k/m or 2πf = √k/m
so you see that straight away you can calculate the frequency of the SHM from the stiffness and mass of the system.

sorry but I believe your explanation is flawed.

the ω in my equation refers to angular frequency, but in circular motion, the ω is angular velocity. They are two different things and cannot be compared as such.

 

[technician]

OK, have you looked in any textbooks to see the similarities between SHM and circular motion?
Good luck in your search for an explanation

 

[serverxeon]

similarity is not a concrete explanation.
it may be coincidence.

 

[maxwellcauchy]

hi, thanks for the respond.

could u please explain how you arrive at

a = d2y/dt2 = v*(dv/dx)?

why is there 'y' and 'x' and why dv/dx is multiplied by v?

sorry,
a = d2y/dt2 = v*(dv/dx) should be 'x' instead of 'y'
So, a = d²x/dt² = v*(dv/dx).
d(dx/dt)/dt = dv/dt = dv/dx * dx/dt = dv/dx * v.

 

[maxwellcauchy]

sorry but I believe your explanation is flawed.

the ω in my equation refers to angular frequency, but in circular motion, the ω is angular velocity. They are two different things and cannot be compared as such.

Simple harmonic motion is the projection of uniform circular motion on a diameter of the circle in which the circular motion occurs.

From Halliday Resnick, Fundamentals of Physics 8E.

That should explain why they are comparable.

 

[Ken G]

similarity is not a concrete explanation.
it may be coincidence.

It's not coincidence, all you have to do is analyze the circular motion one component at a time. In circular motion, you have a force of constant magnitude but changing direction, and when you project such a force onto anyone fixed direction, you will immediately get the force law of simple harmonic motion (try the trig, or just look at the Cartesian coordinates of a force of constant magnitude). Indeed, I find it quite useful to go the opposite direction-- take a 1D simple harmonic oscillator and introduce a second dimension, even if there is really only one. Now imagine the oscillator moves in both dimensions, but just give the motion in the second dimension a 90 degree temporal phase shift (it does everything the 1D solution does, just a quarter period later), thinking of it as a kind of "lagged solution" perpendicular to the first. With a little effort, you can see that the combined motion is in a circle at constant velocity, and circular motion is much easier to solve, because the only thing varying in time is the direction of motion, and you know just how that varies (like a clock). When you've solved the steady state in that situation (even if there is driving and damping and the whole 9 yards), it is a simple matter to remove the "lagged" solution that you originally added, and you recover the 1D solution (by the linearity of the equations).

Note also that springs can really do this kind of motion (you can make a stretched spring go around in a circle), and in fact this is always the easiest device to imagine what springs do. We should imagine that the "natural" thing for a spring to do is go in a circle, and think of 1D oscillation as a special case of that (either project the circular motion, or average it with the circular motion in the opposite direction), rather than the usual way springs are taught as objects that naturally oscillate in 1D. In other words, we understand systems by considering their steady-state response, and the "steady-state" thing that springs do is go in a circle.

 

[technician]

Consider a point in uniform circular motion.
The projection of this point onto a diameter produces a point that is executing SHM
The link could not be stronger.

 

[serverxeon]

ok i get where you all are coming from for the case of comparison.

thanks