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Calculate the period and frequency of a 3.500 m long pendulum at the following locations:

a. the North Pole, where ag= 9.832 m/s^2
b. Chicago, where ag= 9.803 m/s^2
c. Jakarta, Indonesia, where ag= 9.782 m/s^2

User ChuckE
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1 Answer

4 votes

Answer:

1)

The period of a simple pendulum is given by the formula:


T=2\pi \sqrt{(L)/(g)}

where

L is the length of the pendulum

g is the acceleration due to gravity

In this problem we have:

L = 3.50 m is the length of the pendulum

At the North Pole, we have


g=9.832 m/s^2

So the period is


T=2\pi \sqrt{(3.50)/(9.832)}=3.749 s

At Chicago, we have


g=9.803 m/s^2

So the period is


T=2\pi \sqrt{(3.50)/(9.803)}=3.754 s

At Jakarta, we have


g=9.782 m/s^2

So the period is


T=2\pi \sqrt{(3.50)/(9.782)}=3.758 s

2)

The frequency of an object in simple harmonic motion is equal to the reciprocal of the period:


f=(1)/(T)

At the North Pole, we have

T = 3.749 s

So the frequency is


f=(1)/(3.479)=0.267 Hz

At Chicago, we have

T = 3.754 s

So the frequency is


f=(1)/(3.754)=0.266 Hz

At Jakarta, we have

T = 3.758 s

So the frequency is


f=(1)/(3.758)=0.266 Hz

User Elsimer
by
4.0k points