# Simple pendulum equations

1. Dec 16, 2008

### musicfairy

A simple pendulum has a period of 2 s for small amplitude oscillations.

1. The length of the pendulum is most nearly
(A) l/6 m
(B) ¼ m
(C) ½ m
(D) 1 m
(E) 2 m

I used the period of a pendulum equation, did some algebra, and got D as the answer. Is this correct?

2. Which of the following equations could represent the angle θ that the pendulum makes with the vertical as a function of time t?

(A) θ = θmax sin (π/2)t

(B) θ = θmax sin πt

(C) θ = θmax sin 2πt

(D) θ = θmax sin 4πt

(E) θ = θmax sin 8πt

I don't get this one. I'm assuming it's C because 2π = ω in a lot of equations and maybe it's true here too? Can someone please explain this?

2. Dec 17, 2008

### Redbelly98

Staff Emeritus
1. Yes, good.

2. No. One way to think about this is, after one period the argument of the sine should be 2π. And you know what t is after one period.

3. Dec 17, 2008

### musicfairy

I redid #2 this way.

T = 2π/ω
Substituted 2s for T and got ω = π

So the answer would be B?

4. Dec 17, 2008

### Redbelly98

Staff Emeritus
Yes.

5. Apr 27, 2010

### An0maly

Could someone help me out for #2. I'm not even sure how to approach that part.

6. May 9, 2010

7. May 9, 2010

### thebigstar25

They solved it check the previous posts ..

8. May 9, 2010

### An0maly

I saw the answer but I didn't understand why they are plugging in T. What exactly is the concept of θ = θmax sin Wt? I've never learned that so I don't know how to manipulate the equation.

9. May 9, 2010

### thebigstar25

you can reach to the equation θ = θmax sin Wt by considering (i.e.) a pendulum of length l and start from newtons 2nd law which states that net force = m x` , with little work you will end up with the solution θ = θmax sin Wt, where W = sqrt(g/l) ..

you can approach the same answer by substituting l = 1m, g=10m/s^2 in W , then you will get W = 3.16 for g = 10m/s^2 (or W=3.13 for g=9.8 m/s^2) which is approximately equal to the answer they reached .. :)