Question

The earth's acceleration due to gravity varies from

9.780 m/s2 at the equator to 9.832 m/s2 at the poles. A pendulum whose length is precisely 1.000 m can be used to measure g. Such a device is called a gravimeter.

A) How long do 100 oscillations take at the equator?
Express your answer to four significant figures and include the appropriate units.
t= _____?
B) How long do 100 oscillations take at the north pole?
Express your answer to four significant figures and include the appropriate units.
t = _____?
C)

Answer 1

Answer:

A) 200.9s

B) 200.4s

Explanation:

Parameters given:

acceleration due to gravity, g at the equator = $9.780 m/s^{2}$

acceleration due to gravity, g at the poles = $9.832 m/s^{2}$

length of pendulum, l = 1m

number of oscillations = 100

The period of an oscillation is the time taken for a pendulum to complete one oscillation, and it is given as:

$T = 2\pi \sqrt{\frac{l}{g} }$

where l = length of pendulum

g = acceleration due to gravity

A) At the equator, when g = $9.780 m/s^{2}$, the period for one oscillation will be:

$T = 2\pi \sqrt{\frac{1}{9.78} }$

$T = 2\pi * 0.3198\\\\T = 2.009s$

Hence, it will take 2.009s to complete one oscillation. To find the time for 100 oscillations

$T_{100} = 100 * 2.009\\\\T_{100} = 200.9s$

It will take 200.9s for the pendulum to complete 100 oscillations at the equator.

B) At the North pole, when g = $9.832 m/s^{2}$, the period for one oscillation will be:

$T = 2\pi \sqrt{ \frac{1}{9.832}}$

$T = 2\pi * 0.3189\\\\T = 2.004s$

Hence, it will take 2.004s to complete one oscillation. To find the time for 100 oscillations:

$T_{100} = 100 * 2.004\\\\T_{100} = 200.4s$

It will take 200.4s for the pendulum to complete 100 oscillations at the North pole.