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

1. Aug 14, 2010

### mmainak

1. The problem statement, all variables and given/known data

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)}.

2. Relevant 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 ?
3. 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.

2. Aug 14, 2010

### 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.

3. Aug 16, 2010

### 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?

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Last edited: Aug 16, 2010
4. Aug 16, 2010

### 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?

5. Aug 16, 2010

### 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 ?

6. Aug 16, 2010

### diazona

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.

7. Aug 18, 2010

### vela

Staff Emeritus
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.