Answer:
1)
The period of a simple pendulum is given by the formula:
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
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
[tex]g=9.832 m/s^2[/tex]
So the period is
[tex]T=2\pi \sqrt{\frac{3.50}{9.832}}=3.749 s[/tex]
At Chicago, we have
[tex]g=9.803 m/s^2[/tex]
So the period is
[tex]T=2\pi \sqrt{\frac{3.50}{9.803}}=3.754 s[/tex]
At Jakarta, we have
[tex]g=9.782 m/s^2[/tex]
So the period is
[tex]T=2\pi \sqrt{\frac{3.50}{9.782}}=3.758 s[/tex]
2)
The frequency of an object in simple harmonic motion is equal to the reciprocal of the period:
[tex]f=\frac{1}{T}[/tex]
At the North Pole, we have
T = 3.749 s
So the frequency is
[tex]f=\frac{1}{3.479}=0.267 Hz[/tex]
At Chicago, we have
T = 3.754 s
So the frequency is
[tex]f=\frac{1}{3.754}=0.266 Hz[/tex]
At Jakarta, we have
T = 3.758 s
So the frequency is
[tex]f=\frac{1}{3.758}=0.266 Hz[/tex]
