# Amplitude in pendulum motion

1. Nov 18, 2007

### gills

Amplitude in pendulum motion....

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

2. Relevant equations

w = $$\frac{g}{L}$$

T = 2$$\pi$$$$\sqrt{\frac{L}{g}}$$

I$$\left(\frac{d^{2}\theta}{dt^{2}}\right)$$ = -mgL$$\theta$$

x(t) = Acos(wt) ??

3. The attempt at a solution

I'm stuck on this problem. I haven't worked with harmonic motion enough yet to grasp it. Any help would be greatly appreciated.

I need a hand on this.

2. Nov 18, 2007

### rl.bhat

At the lowermost position of the mass of the pendulum the tension on the string is equal to the centripetal reaction. i.e. T= mv^2/r. You can find the velocity by using the conservation of energy. i.e. K.E = P.E.

3. Nov 18, 2007

### Shooting Star

So, where are you putting the weight of the mass at the lowermost position?

4. Nov 18, 2007

### rl.bhat

In the pendulum the mass is swinging. During the swing when the mass crosses the lowermost point, the tension in string is maximum and it is equal to mv^2/r + mg According to the conservation of energy 1/2*m*v^2 = mgh, where h is the height from mthe lower point from which the mass of the pendulum is released. If you know you can find the amplitude.

Last edited: Nov 18, 2007
5. Nov 18, 2007

### Staff: Mentor

Incorrect. There are two forces acting on the mass at that point; tension is one of them. (This was Shooting star's point.) The net force equals mv^2/r.

Right!

6. Nov 18, 2007

### rl.bhat

Right. Tension = mv^2/r + mg.

7. Nov 18, 2007

### gills

wow, so it turns out that we don't need to use any concepts from this particular chapter. I was actually onto this the way you guys described it, but then i said to myself that we must have to use some formula or something from this chapter. I guess not.

T = $$\frac{mv^{2}}{r}$$ + mg --->

v = $$\sqrt{r\left(\frac{T}{m} - g\right)}$$

v = 1 m/s --->

E$$_{i}$$ = E$$_{f}$$

mgh = $$\frac{1}{2}$$mv$$^{2}$$ --->

h = $$\frac{v^{2}}{2g}$$ --->

h = 0.05m --->

then from my free body diagram

0.45 - 0.05 = 0.40m

cos$$^{-1}$$$$\left(\frac{0.40}{0.45}\right)$$ = 27.27$$^{o}$$

The textbook answer is 27.16$$^{0}$$. I always get answers that are a little off from the text. Calculators?