# Length of a String - Pendulum

1. Jan 12, 2009

### hellokitty

1. The problem statement, all variables and given/known data
A 125 g ball is tied to a string. It is pulled to an angle of 4.80 degrees and released to swing as a pendulum. 12 oscillations take 9.00 s.

2. Jan 12, 2009

### Staff: Mentor

What do you know about pendulums? What's the formula for the period of a pendulum?

3. Jan 12, 2009

### hellokitty

Period (T) = 1/f
T = 2π √(l/g)

4. Jan 12, 2009

### Staff: Mentor

That's true, but that's true in general, not just for pendulums. What formula allows you to calculate the period of a pendulum from its physical properties, such as its length?

That's the one.

Hint: Use the data provided to find the period. Then solve for the unknown length.

5. Jan 12, 2009

### hellokitty

I did that, but I wasn't sure if I got the correct answer.

T = 12/9 = 1.33333

1.3333=2π√(L/9.8)
1.3333(2π)= √(L/9.8)
squared the left side and multiplied it by 9.8 and..

I eventually got 687.07m

6. Jan 12, 2009

### Staff: Mentor

Careful. The period is the time for one oscillation, so it should be total time divided by the number of oscillations. (You have them switched.)

7. Jan 12, 2009

### hellokitty

ah, I see.

Thanks!

I haven't had any practice in physics for a long time, so I'm trying to re-learn everything.

8. Jan 12, 2009

### hellokitty

I just realized that I didn't use the mass or degree of angle in solving the question.

Is there a way to answer this question using another equation?

9. Jan 12, 2009

### Staff: Mentor

Perhaps the period does not depend on either of those variables.

Are there are more parts to this question?

10. Jan 12, 2009

### hellokitty

no, this was the entire question.

I guess the mass and angle is there to play mind tricks.

11. Jan 12, 2009

### hellokitty

Another question:

Astronauts on the first trip to Mars take along a pendulum that has a period on earth of 1.50 s. The period on Mars turns out to be 2.45 s.

I used the equation T=2pi √ (L/g)
I first plugged in the 1.5s = 2pi √ (L/9.8) and found L = .5585

Then I plugged in 2.45s = 2pi √(.5585/g) to solve for g of mars.

Now I got 3.6733 m/s2 = g for mars.

Can someone correct me if I'm wrong or if there is an easier way to solve this problem?

12. Jan 13, 2009

### Staff: Mentor

Yours is a perfectly fine way to solve this problem.

Here's how I would solve it:

$$T_e = 2\pi \sqrt{L/g_e}$$

$$T_m = 2\pi \sqrt{L/g_m}$$

Divide the two and square:
$$(T_e/T_m)^2 = g_m/g_e$$

Thus:
$$g_m = g_e(T_e/T_m)^2$$