Differential equation for a pendulum

[Faiq]

Homework Statement

A simple pendulum is formed by a light string of length ##l## and with a small bob ##B## of mass ##m## at one end. The strings hang from a fixed point at another end. The string makes an angle ##\theta## with the vertical at time ##t##. Write down an equation of motion of ##B## perpendicular to ##OB## and state the approximation that enables you to treat this equation as an SHM equation. State the general solution of this equation.

The Attempt at a Solution

Equation of motion : ## \rm \large -mgl\sin(\pi -\theta) = ml^2 \ddot \theta##
Rearranges to: ## \rm \large \ddot \theta =-\frac{g}{l}sin(\pi-\theta)##
Approximation: ## \rm \large sin \theta \approx \theta## provided ## \theta ## is small.
General solution: ## \rm \large \theta = \theta_0sin(\sqrt{\frac{g}{l}}t)##

Are my answers correct?
The reason I am confused is because I think I have misunderstood what they mean by "equation of motion of ##B## perpendicular to ##OB##" and may have got all of my answers wrong.
And secondly I am well aware of finding a general solution of a differential equation of the form ## \rm \small \ddot y =ay ##, however, I have never seen a general solution of a differential equation of the form ##
\rm \small \ddot y =a(\lambda-y) ## where ##\lambda## is a constant.

 

[Dr.D]

It is not entirely clear what you are calling theta (and thus why you have involved a pi - theta argument for the sine). A figure would really help your question, along with labels on the figure.

As to the solution for the differential equation, did you really mean ddy = a*y, where a is presumed to be positive? Or did you intend to have it be negative?

The additional term a*lambda simply adds a constant to the particular solution.

 

[Chestermiller]

Why are you interested in the solution to that differential equation?

 

[Faiq]

It is not entirely clear what you are calling theta (and thus why you have involved a pi - theta argument for the sine). A figure would really help your question, along with labels on the figure.

I hope this will work.

As to the solution for the differential equation, did you really mean ddy = a*y, where a is presumed to be positive? Or did you intend to have it be negative?

The additional term a*lambda simply adds a constant to the particular solution.

Yes, can you show me how to add the constant to the particular solution? Like for example if the solution to the differential equation ## \ddot y = ay## where a is a constant is ## y = y_0 \sin \omega t##, how will the constant term be added in the solution if the equation is changed to ## \ddot y = a(\lambda-y)##? (If changes vary from equation to equation, please refer to the SHM equation given in my first post) If you find my example vague, feel free to make one of yours.

Why are you interested in the solution to that differential equation?

I am not interested in that particular equation. I am just concerned whether changes will occur in the solution if the concerned variable was increased or decreased by a constant term.

 

[Dr.D]

Start by replacing sin(pi-theta) with just sin(theta); they are exactly equal.

Then, with a little bit of re-arrangement, your equation should look like
dd(theta) + w^2*sin(theta) = 0

Then make the approximation that for small theta, sin(theta) = theta, so you have
dd(theta) + w^2*theta = 0

That should be the only equation you need to solve for this problem.
I really don't know where the question about a constant term on the right came from.

 

[Faiq]

Okay thank you that explains a lot.