Homework Help: 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.

Please help, I'm not sure which formulas to use

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.

My answer:

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

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