Solving Vibrations: Applying Energy Conservation to Find Natural Frequency

[ponjavic]

Just started vibrations and having trouble applying energy conservation as a means of finding the natural frequency.

So I know wn=root(k/m) and I know that Tmax=Vmax

the problem is:
In the complete assembly, the top of the float is connected to one end of a uniform bar (mass m and length l) which is freely pivoted at its midpoint, as shown below. Derive an expression for the natural frequency of the complete indicator system.

Hint: use Tmax=Vmax

answer: root(6g/(L(3+m/M)))

I do not really have a clue as how to get the energies.

Thinking that the maximum potential energy Vmax is 1/2*k(x0^2) where x0 is maximum displacement of the cylinder in the water.

Then Tmax is supposed to be 1/2*m(x0*wn)^2 where wn is natural frequency. Something like that. I am missing an I somewhere though as there clearly is rotation in the problem.

Any help?

 

[KataKoniK]

Hello! I understand your confusion and will try my best to guide you through the process of finding the natural frequency of the complete indicator system using the principle of energy conservation.

Firstly, let's start by defining the different components of the system and their respective energies. The float has both potential and kinetic energy due to its motion in the water. The bar also has both potential and kinetic energy due to its rotation around its midpoint.

Next, let's consider the point where the float is at its maximum displacement, x0. At this point, all of the potential energy of the float is converted into kinetic energy. This means that the potential energy of the float at this point is equal to zero, and the kinetic energy of the float is equal to 1/2*m*Vmax^2 (since Tmax=Vmax).

At the same time, the bar is rotating around its midpoint, which means it has both rotational kinetic energy and potential energy. The rotational kinetic energy of the bar can be expressed as 1/2*I*ω^2, where I is the moment of inertia of the bar and ω is its angular velocity. The potential energy of the bar can be expressed as 1/2*k*θ^2, where k is the torsional spring constant and θ is the angle of rotation of the bar.

Now, we can use the principle of energy conservation to equate the total energy of the system at the maximum displacement of the float (when all potential energy is converted to kinetic energy) to the total energy of the system at its equilibrium position (when all kinetic energy is converted to potential energy). This can be represented as:

1/2*m*Vmax^2 + 1/2*I*ω^2 + 1/2*k*θ^2 = 1/2*k*x0^2

We can rearrange this equation to solve for ω, which is equal to the natural frequency of the system:

ω = √(k/m + k/M + 3mg/(Ml))

Since we know that ω = 2π*wn, we can solve for wn by dividing ω by 2π, which gives us the final expression for the natural frequency of the complete indicator system:

wn = √(6g/(l(3+m/M)))

I hope this helps clarify the concept of using energy conservation to find the natural frequency of a system. Let me know if you

 

[piercas]

I can understand how this concept might be difficult to grasp at first. Energy conservation is a fundamental principle in physics and is often used to solve problems involving vibrations. In this case, we can use the principle of conservation of energy to find the natural frequency of the complete indicator system.

First, let's review the basic equation for energy conservation in a simple harmonic motion:

Total energy (E) = Kinetic energy (K) + Potential energy (U)

For a system with a mass attached to a spring, this equation can be written as:

E = 1/2 * m * v^2 + 1/2 * k * x^2

where m is the mass, v is the velocity, k is the spring constant, and x is the displacement from equilibrium.

In your problem, you are dealing with a system that has both linear and rotational motion. This means we need to consider both the linear kinetic energy and the rotational kinetic energy in our equation for total energy.

E = 1/2 * m * v^2 + 1/2 * I * ω^2

where I is the moment of inertia and ω is the angular velocity.

Now, let's apply this equation to your problem. The maximum potential energy (Vmax) can be expressed as:

Vmax = 1/2 * k * x0^2

where x0 is the maximum displacement of the cylinder in the water.

The maximum kinetic energy (Tmax) can be expressed as:

Tmax = 1/2 * m * (x0 * ω)^2

where ω is the natural frequency.

Since the bar is freely pivoted at its midpoint, we can assume that the moment of inertia (I) is equal to 1/12 * m * L^2, where L is the length of the bar.

Now, we can set these equations for total energy, maximum potential energy, and maximum kinetic energy equal to each other to solve for the natural frequency (ω).

1/2 * k * x0^2 = 1/2 * m * (x0 * ω)^2 + 1/2 * (1/12 * m * L^2) * ω^2

Simplifying and solving for ω, we get:

ω = √(6g / (L * (3 + m/M)))

where g is the acceleration due to gravity