Frequency of a spring in an extended system

[Violagirl]

Homework Statement

Beginning with
basic physics principles, show how you get an equation that gives the frequency of
oscillation.

Homework Equations

Newton second law of motion: Force = mass x acceleration

Hooke's law for the force of the spring: F(spring) = -k(constant)x X(displacement of spring)

Forces of system:

In x direction: T-kx=ma

In y direction: T-mg=ma=0, T=mg

Total forces of system: mg-kx=ma

In measuring angular frequency: ω = √k/m

For measuring frequency: f = 1/2∏√k/m

For measuring period: T = 2∏√m/k

The Attempt at a Solution

Starting Newton's Second Law of Motion:

∑F=ma

In the +x direction to the attached free body diagram of the system, we have the force of the spring:

Fapplied = kx

Since the spring will be displaced from it's point of equilibrium to be released, I can show this by the equation:

∑Fx=mxax

Fspring=-kx

When an amount of mass (m) is attached to the end of a spring, and displaced from equilibrium to be released, the unbalanced force acting on the mass that remains is the force exerted by spring. Through Newton's Second Law of Motion, this can be expressed by the equation shown below:

Fnet= Fspring =-kx=ma

Showing the differential equation,

m(d2x/dt2)=-kx

we can get the simple harmonic equation:

x=A cos(ωt- Θ)

After showing that the harmonic and differential equations are equivalent through direct substitution, we can get the equation for finding angular frequency:

ω=√k/m

The motion of the spring's oscialliations can be shown by it's period (T) as a unit of time. Because it's motion depicts the cycle of a circle, a diameter of 2ττ, we can get an equation of:

T = 1/2ττ √k/m

Taking the reciprocal, we can get the frequency:

f=1/T= 2ττ √m/k

Since the mass of both the cart and hanging object impact the frequency of the spring, I believe the following equation would correspond to it:

f = 2ττ √m1+m2/k

Is this wrong?

 

[Violagirl]

Free body diagram of extended system is attached.