# Problem in harmonic motion

In the harmonic chapter of my book it is written that $$\omega^{2}=\frac{k}{m}$$ where $$\omega$$ is frequency and $$k$$ is the spring constant. But how did they get to this formula? What is its derivation?

I read somewhere that there is an analogy between the motion of a mass on a spring and a mass going in circles in a certain velocity. How can they be analogous? I mean aren't they totally different?

It is enough to verify that the proposed solution actually solves the differential equation but to get some insight, start with the equation of motion $$\frac{d^2x}{dt^2}=-\frac{k}{m}x \tag{1}$$ ask yourself: what functions have the property that its second derivative is the negative of itself, up to a multiple? (This is, after all, what equation (1) says.) The answers are the sine and cosine function since, for instance, $$\frac{d^2}{dt^2}A\cos\omega t=-\omega^2 A\cos\omega t.$$ Comparing this with (1) yields $$x=A\cos\omega t$$ and $$\omega=\pm \sqrt{k/m}$$. Since $$e^{i\omega t}=\cos\omega t+i\sin\omega t$$, the solutions to (1) can also be written in terms of $$e^{i\omega t}$$ and $$e^{-i\omega t}$$, and in particular $$\frac{d^2}{dt^2} e^{\pm i\omega t}=-\omega^2 e^{\pm i\omega t}$$ so these are also solutions, albeit complex functions.

As to the connection with circular motion: a particle with constant angular velocity $$\omega$$ travelling on a circle of radius $$A$$ has coordinates given by $$x= A\cos\omega t\, ,\qquad y=A \sin\omega t\, ,\qquad x^2+y^2=A^2$$ where $$\omega=\frac{d\theta}{dt}$$ is the angular velocity. Thus, the projection of this circular 2d motion on the $$x$$ or $$y$$ axis, or alternatively the $$x$$ or $$y$$ part of the motion separately, have the same form as the motion of a particle on a harmonic oscillator.

This the angular frequency, which comes from solving the laplacian equation of $$\frac{d^2x}{dt^2} = -\frac{k}{m} x = -\omega^2 x.$$ The solution to this equation is $$A\exp(i\omega x)+B\exp(-i\omega x),$$ where $$\omega$$ is given as above.

Also, real part of this solution is $$A\sin(\omega x+\phi)$$, which is nothing but the projection of a body moving in circular motion on its x-axis. This is how the term angular frequency comes.

• But how did you show the derivation? I didn't see any. – Theoretical Oct 30 '18 at 11:06
• @AsifIqubal He stated the equation of motion á la Newton and gave the general solution. He also related the solution to the circular motion system, as you asked. Where did you find a problem with his answer, maybe then you can have more precise help. – Alejandro Menaya Oct 30 '18 at 12:07
• @AsifIqubal nobody is going to write the contents of a differential equations course and a dynamics course in an answer here. There are hundreds of textbooks and web sites which answer your question in as much detail as you want. – alephzero Oct 30 '18 at 12:22
• @alephzero How can anyone prove the a circular motion is a simple harmonic motion? – Theoretical Oct 30 '18 at 13:19

I read somewhere that there is an analogy between the motion of a mass on a spring and a mass going in circles in a certain velocity. How can they be analogous? I mean aren't they totally different?

In many books, the representative diagram of a uniform circular motion is used to represent an oscillatory system.

the spring attached to a mass m and if it is extended by a displacement say x then the spring force acts opposite to the displacement and pulls the body back.

however, this force generates a kinetic energy of the body and if it comes back to its original length the spring force is zero but the velocity pushes the spring compressing it by a length x and there its kinetic energy becomes zero and spring gets a potential energy to push back the mass and an oscillatory motion is set up.

IF the spring stretches to a maximum value x =A then this becomes the amplitude of oscillation.

If no dissipative forces are present and the spring is massless/ideal the motion is a simple harmonic one. The Force F = -k.x where k is the spring constant which can be written as

mass.acceleration =- k.x

or a = -(k/m).x = -w^2.x

where x is instantaneous displacement of the mass m from mean position.

The acceleration a =0 at x=0 and is maximum at x= +A or -A (A being the amplitude)

the velocity is max. at x=0 and zero at x= +A or -A

this motion can be represented on a diagram - draw a circle of radius A the amplitude and move a radius vector from x=A to x= 0

then from x=0 to x=-A drop a perpendicular from the tip to the x-axis, say at an angle theta from the x-axis. The projection will be A.Cos (theta)

as the tip moves from x-axis to y-axis the projection will move from x=A to x=0 and again to x=-A. Imagine a mass m moving at the tip of the projection then one can see that the mass is oscillating between +A and -A as the radius vector travel the full circle.

Therefore the length of the base, i.e position of mass m can be written as

x = A Cos(wt) if the instantaneous position of the radius vector can be written as theta = w.t where w can be represented as the angular velocity of the radius vector and in a time T it completes one full rotation of 2.pi

w = (2.pi) /T

and each quadrant represents a time span of T/4

and the half circle represents T/2.

Thus the mass m which is placed at the base of the perpendicular oscillates between A and -A during a complete time period of T which gets related to w.

As the force is proportional to displacement in the opposite direction- when the mass reaches A or -A the force is max. i.e. the acceleration is max. but the velocity is momentarily zero and the velocity is maximum when it passes through x=0.

the above picture is a way to represent harmonic motion on a circular diagram. It is actually composed of two SHM's of identical time period but perpendicular to each other with a phase difference pi/2. As one can draw a perpendicular on the y-axis and again the base of the perpendicular will oscillate along the y-axis with y = A Sin (wt) and a mass can be imagined to oscillate along y with similar amplitude and w.