How Does a Rod Attached to a Spring Undergo Simple Harmonic Motion?

[K.QMUL]

Homework Statement

The figure shows a thin uniform rod of mass M and length 2L that is pivoted without friction about an axis through its midpoint.A horizontal light spring ofspring constant k is attached to the lower end of the rod.The spring is at its equilibrium length when the angleθ with respect to the vertical is zero.Show that for oscillations of small amplitude, the rod will undergo SHM with a period of 2π√M/3k. The moment of inertia of the rod about its midpoint is ML2/3.(Assume the small angle approximations)

The Attempt at a Solution

My book gives the following answer but is not indepth, could someone help me out by going through step by step how they came to this answer...

Couple acting on rod = −kLsinθ×Lcosθ=−kL2θ for smallθ.
Hence, I(d^2θ/dt^2)=−kL^2θ

 

[BvU]

Sure. My telepathic powers are limited, however, so I can't see the figure clearly. Is the spring dangling loose or is it attached to something fixed ?

As I said, limited powers, so I can't read what you intended to fill in under 2., either. Could you oblige ?

Then: the english word for couple is torque.

 

[K.QMUL]

I think I've solved it, never mind, thanks for your offer

 

[dauto]

You should also learn to use superscripts. L2 and L² are not the same thing.

 

[paranormal]

or (d^2θ/dt^2)+(kL^2/I)θ=0
Hence ω=√kL^2/I
i.e. T=2π√I/kL^2=2π√M/3k

I would like to first clarify the assumptions made in this problem. The small angle approximation assumes that the angle θ is small enough that the sine and cosine functions can be approximated by their linear terms. This allows us to simplify the equations and make them easier to solve.

Now, let's break down the solution step by step:

1. The first step is to find the couple (torque) acting on the rod. This is done by using the formula for torque, τ = r x F, where r is the distance from the point of rotation to the point where the force is applied, and F is the force. In this case, the force is the tension in the spring, and the distance is Lsinθ (the vertical component of the distance from the pivot to the spring).

2. Next, we use the small angle approximation to simplify the equation and get rid of the sine and cosine terms. This allows us to rewrite the torque as -kL2θ.

3. Using Newton's second law, F=ma, we can write the equation of motion for the oscillations as Ma = -kL2θ, where M is the mass of the rod and a is the acceleration.

4. We can rewrite this equation as a differential equation, (d2θ/dt2) + (kL2/I)θ = 0, where I is the moment of inertia of the rod about its midpoint, ML2/3.

5. Solving this differential equation, we get the angular frequency ω = √kL2/I. This is the frequency at which the rod will oscillate.

6. Finally, we can use the formula for the period of a simple harmonic oscillator, T = 2π/ω, to get the period of oscillation as T = 2π√I/kL2 = 2π√M/3k, using the given values for I and L.

I hope this explanation helps to clarify the steps taken to solve this problem. It is important to carefully consider the assumptions made and to use the correct equations and formulas to arrive at the correct solution.