How Long Does a Pendulum Take to Complete 12 Oscillations?

[hellokitty]

Homework Statement

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

 

[Doc Al]

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

 

[hellokitty]

Period (T) = 1/f
T = 2π √(l/g)

 

[Doc Al]

Period (T) = 1/f

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?

T = 2π √(l/g)

That's the one.

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

 

[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

 

[Doc Al]

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

T = 12/9 = 1.33333

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.)

 

[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.

 

[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?

 

[Doc Al]

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

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

Are there are more parts to this question?

 

[hellokitty]

no, this was the entire question.

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

 

[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?

 

[Doc Al]

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