Is the angular span of a body executing s.h.m always 2.pi?

[mmainak]

Homework Statement

a wall is inclined to an angle "a" to the horizontal plane. a pendulum of length "l" attached to the wall making an angle "b" with the horizontal (a >b)is let to oscillate. Collision between the bob and the wall is perfectly elastic.show that the time period of the oscillation is 2.sqrt{l/g}.{pi/2 + sin[-1](b/a)}.

Homework Equations

In solving this problem i have found that all other terms are same for a pendulum in normal condition except in place of 2.pi there is 2.{pi/2 + sin[-1](b/a)}.So my question is what is the basis that we generally take the angular span of s.h.m to be 2 pi? Can there be any exception ?

The Attempt at a Solution

I tried to solve the problem following way.
from it's hanging point to the mean position it goes an angular distance pi/2.
from mean position to the angle a it goes an angular distance sin[-1](b/a), i can prove it as following.

if the angular distance is theta then
[l.sin a].sin theta = l.sin b
taking a,b small,

sin theta = l.b/l.a
hence, theta = sin[-1](b/a)

So effective angular span =2.{pi/2 + sin[-1](b/a)}
so time period = 2.sqrt{l/g}.{pi/2 + sin[-1](b/a)}.

Please tell me whether I am right in my concept or not.

 

[diazona]

Simple harmonic oscillators execute sinusoidal oscillations, by definition. (That's why they're called "harmonic") The period of the sine function is 2π, so yes, a true SHO will always go through a rotation of 2π per cycle in "parameter space". What you have here is not a true SHO, though - it's more like a truncated SHO.

There are a couple of things I don't understand about your proof, though (maybe I'm just missing something obvious): how exactly do you get this equation?
$$l\sin a \sin\theta = l\sin b$$
Also, how can you assume that a and b are small? Those are angles from the horizontal axis, but in order to treat a pendulum as an SHO, it must stay close to the vertical axis.

If you have a diagram you can post, I'm sure that would be helpful.

 

[mmainak]

Here I have uploaded two diagram of this problem. here I have approximated a,b being small
r= l.a
r sin$$\vartheta$$=l.b

hence $$\vartheta$$ =sin ^-1b/a

well I also feel that I have over-approxiamted this problem . Is there any other way I can proceed?

 

[diazona]

Your two diagrams are inconsistent - in diagram 1, the angles are defined relative to the horizontal axis (as in the problem statement), but in diagram 2, they're defined relative to the vertical axis. Which is it?

 

[mmainak]

yes that was mistaken . sorry

dia 1 is the original one given in the problem

instead of l.a it should be l.(pi/2 -a)

but that will not make the answer. how can i proceed then ?

 

[diazona]

yes that was mistaken . sorry

dia 1 is the original one given in the problem

instead of l.a it should be l.(pi/2 -a)

but that will not make the answer. how can i proceed then ?

Honestly, I can't figure out how to get the answer you're supposed to get. Perhaps someone else can come along and offer a second opinion.

 

[vela]

The expression you're asked to derive is incorrect. When the wall is vertical, the period of oscillation should be half the usual period, i.e. $$T=\pi\sqrt{l/g}$$. That doesn't match what the expression predicts.