# Simple pendulum swing problem

1. Apr 1, 2004

### lollypop

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 47.0 cm. She finds that the pendulum makes 97.0 complete swings in a time of 141 s .
What is the value of g on this planet?

a complete swing is a cycle??? so the period would be 97/ 141 = 0.688??is this right?
i was thinking of using the formula frequency = (1/2pi)*sqrt(g/ L), and get g from there. I tried using the period i said before but it is not right. What am i doing wrong?

2. Apr 1, 2004

Post your work. Your approach is correct.

3. Apr 1, 2004

### lollypop

so g = (f*2pi)^2 * L

f = 1/period = 1.453 Hz

which gives me g= 39.2 and this is wrong .

4. Apr 1, 2004

Check your algebra. I'm getting a different answer.

5. Apr 1, 2004

### lollypop

i have checked the algebra and i keep on getting 39.2 m/s^2 as the gravity, is my formula right???

6. Apr 1, 2004

T = 2pi*Sqrt[L/g]

g = L*T^2/(4pi^2)

Yes?

7. Apr 1, 2004

### lollypop

by your formula u get 0.0082????? how did u get T^2 on the numerator, when I solve for g using your formula I get (L*4pi^2)/(T^2) which i think is right.

8. Apr 1, 2004

Eh, heh, heh... I suck at typing.

g = 4pi^2*L/T^2

9. Apr 1, 2004

### lollypop

which gives 39.2 for g. what other number u get that u said was different??

10. Apr 1, 2004

4pi^2*(.47m)/(141s/97)^2 = ?

11. Apr 2, 2004

### HallsofIvy

Staff Emeritus
You have an error back in your first post:

"a complete swing is a cycle??? so the period would be 97/ 141 = 0.688??is this right? "

No, it's not right. It would be a good idea to carry the units along with your calculation. The problem tells you that the pendulum "makes 97.0 complete swings in a time of 141 s ." so that 97/141 would be 97/141 swings per second which is not "period". It is, in fact, the frequency in cycles per second.

The frequency is 0.688 cycles per second, not the period.

12. Apr 2, 2004

### Chen

(And the period is one over frequency.)