Simple Harmonic Motion of a plank of mass

[Mishy]

I don't understand the solution to the following problem:

A horizontal plank of mass m and length L is pivoted at one end. The plank's other end is supported by a spring of force constant k. The moment of inertia of the plank about the pivot is (1/3)m(L^2). The plank is displaced by a small angle theta from its horizontal equilibrium position. Show that it moves with simple harmonic motion with angular frequency omega = square root (3k/m).

Here is my work:
net torque = I[I]*alpha [/I]
net torque = (1/3)m(L^2)*alpha
kxd = (1/3)m(L^2)*(second derivative of theta)
x is the vertical distance from the horizontal equilibrium position.
kxLsin(theta) = (1/3)m(L^2)*(second derivative of theta)
kxL(theta) / (mL^2) = second derivative of theta
because theta is small, sin(theta) = theta
kx(theta)/ mL = second derivative of theta
using this, I would get omega as the square root of (3k/mL)
How do I get rid of that L in my formula for omega? Where was I supposed to drop it?

Thanks.

 

[Physics Monkey]

Your formula for angular velocity can't be right because it doesn't have the right units to be an angular velocity. So what did you do wrong?

You have this equation $$\frac{d^2\theta}{dt^2} = \frac{3 k }{m L} x$$, but you can't just conclude that everything in front of the x is equal to $$\omega^2$$. To see why not, think about the units. What you need to do is relate x to $$\theta$$, then try to extract the angular frequency.

Hope this helps.

 

[kyle8921]

I can provide an explanation and solution to your problem. First, let's start by understanding what simple harmonic motion is. It is a type of periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts in the opposite direction of the displacement. This results in a sinusoidal motion with a constant amplitude and frequency.

In this problem, we have a horizontal plank that is pivoted at one end and supported by a spring at the other end. The plank is displaced by a small angle theta from its horizontal equilibrium position. To show that it moves with simple harmonic motion, we need to prove that the restoring force is directly proportional to the displacement and acts in the opposite direction.

Let's start by looking at the forces acting on the plank. We have the weight of the plank acting downwards, the normal force from the pivot acting upwards, and the spring force acting in the opposite direction of the displacement. Since the plank is horizontal, we can ignore the weight and normal force and focus on the spring force.

The spring force can be represented by Hooke's Law, F = -kx, where k is the force constant and x is the displacement from the equilibrium position. In this case, x is the vertical distance from the horizontal equilibrium position. Now, let's look at the moment of inertia of the plank about the pivot. It is given by (1/3)m(L^2), where m is the mass of the plank and L is its length.

Using Newton's Second Law for rotational motion, we can write the equation for the net torque acting on the plank as I*alpha, where I is the moment of inertia and alpha is the angular acceleration. Substituting the values, we get:

kxLsin(theta) = (1/3)m(L^2)*alpha

Now, for small angles, we can approximate sin(theta) as theta. This is known as the small angle approximation. Using this, we get:

kxL(theta) = (1/3)m(L^2)*alpha

We also know that alpha is equal to the second derivative of theta with respect to time. So, we can rewrite the equation as:

kxL(theta) = (1/3)m(L^2)*(second derivative of theta)

Now, we need to solve for the angular frequency, omega. To do this, we need to get rid of the L in the equation. We can do this