Deriving angular frequency for simple harmonic motion

In summary: Therefore, the equation for angular frequency (\omega) for simple harmonic motion of a spring is \omega = \sqrt{\frac{k}{m}}.
  • #1
AlexYH
1
0

Homework Statement


Derive the equation for angular frequency for simple harmonic motion of a spring.


Homework Equations


Derive omega = sqrt(k/m) from F = -kx
(sorry i don't know how to use notation)


The Attempt at a Solution


I asked my teacher how to do this, and he used some crazy math I didn't learn yet, including Euler's identity and differential equations. I'm in an AP calculus bc class, and i understand differential equations, just not some aspects. Does anyone know a simple solution for this? Thanks in advance
 
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  • #2
If [tex]F=ma[/tex] and [tex]F=-kx[/tex]

Then [tex]ma=-kx[/tex] (by equating the forces.)

Which can be also written as [tex]ma+kx=0[/tex]

or [tex]a+\frac{k}{m}x=0[/tex]

Now if x is displacement, differentiating once with respect to time will give you velocity of the spring and then differentiating again with respect to time will give acceleration.

Displacement of a spring can be given by

[tex]x=A * Cos (\omega t)[/tex]

where A is the Amplitude of motion and [tex]\omega [/tex] is the angular frequency

Now Differenting once will give velocity;

[tex]v=-A\omega Sin(\omega t)[/tex]

and again to give acceleration

[tex]a=-A \omega^{2} Cos(\omega t)[/tex]

Now substituting our formula for Acceleration and displacement into our equation of motion

[tex]a+\frac{k}{m}x=0[/tex]

Gives [tex]-A \omega^{2} Cos(\omega t) +\frac{k}{m}A Cos (\omega t)=0[/tex]

Which can be rearranged to;

[tex]A(-\omega^{2} +\frac{k}{m})Cos(\omega t)=0[/tex]

Can get rid of the [tex]A[/tex] and [tex]Cos(\omega t)[/tex]

which leaves [tex]-\omega^{2} +\frac{k}{m}=0[/tex]

which can be rearranged to [tex]\omega=\sqrt{\frac{k}{m}}[/tex]
 
  • #3
tjr39 said:
If [tex]F=ma[/tex] and [tex]F=-kx[/tex]

Then [tex]ma=-kx[/tex] (by equating the forces.)

Which can be also written as [tex]ma+kx=0[/tex]

or [tex]a+\frac{k}{m}x=0[/tex]

or from here

that is in the form [itex]a=-\omega^2x[/itex]

where [itex]\omega^2=\frac{k}{m}[/itex]
 

1. What is angular frequency in simple harmonic motion?

Angular frequency is a measure of the rate at which an object oscillates back and forth around a fixed point in a circular motion. It is represented by the symbol ω (omega) and is measured in radians per second.

2. How do you derive the angular frequency for simple harmonic motion?

To derive the angular frequency for simple harmonic motion, you can use the equation ω = 2π/T, where T is the period of the motion. This equation states that the angular frequency is equal to 2π divided by the period, which is the time it takes for one complete cycle of the motion.

3. What is the relationship between angular frequency and frequency in simple harmonic motion?

The relationship between angular frequency and frequency in simple harmonic motion is that they are inversely proportional. This means that as the angular frequency increases, the frequency decreases, and vice versa. The equation for this relationship is ω = 2πf, where f is the frequency in hertz.

4. How does changing the mass or spring constant affect the angular frequency in simple harmonic motion?

Changing the mass or spring constant in simple harmonic motion affects the angular frequency by altering the period of the motion. The larger the mass or spring constant, the longer the period and the smaller the angular frequency. This is because a larger mass or spring constant requires more energy to complete one cycle of the motion.

5. Can the angular frequency of simple harmonic motion be negative?

No, the angular frequency of simple harmonic motion cannot be negative. It is a measure of the rate of change, and since oscillations occur in a positive direction, the angular frequency must also be positive. A negative angular frequency would indicate a motion in the opposite direction, which is not possible in simple harmonic motion.

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