# Simple Harmonic Motion: Pendulum

• TJC747
In summary, the Earth's acceleration due to gravity varies from 9.78 m/s2 at the equator to 9.83 m/s2 at the poles. A pendulum with a length of 1.000 m, known as a gravimeter, can be used to measure g. To find the time for 100 oscillations at the equator, use the formula T= 2*pi*sqrt(L/g). Similarly at the North Pole, use the same formula. If the gravimeter is taken to a high mountain peak near the equator and 100 oscillations take 201 seconds, g can be calculated using the same formula.
TJC747
The Earth's acceleration due the gravity varies from 9.78 m/s2 at the equator to 9.83 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?
(in sec)

(b) How long do 100 oscillations take at the north pole?
(in sec)

(b) Suppose you take your gravimeter to the top of a high mountain peak near the equator. There you find that 100 oscillations take 201 seconds. What is g on the mountain top?
( in m/s^2)

I guess I'd use T = 2*pi*sqrt(L/g) in conjunction with other Hooke formulae. Help would be appreciated. Thanks.

right, so use

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

to find the time for one oscillation

(a) At the equator, the acceleration due to gravity is 9.78 m/s^2. Using the equation T = 2*pi*sqrt(L/g), where T is the period of oscillation and L is the length of the pendulum, we can calculate the time for 100 oscillations: T = 2*pi*sqrt(1.000/9.78) = 2*pi*0.324 = 2.03 seconds. Therefore, 100 oscillations would take 2.03*100 = 203 seconds at the equator.

(b) At the north pole, the acceleration due to gravity is 9.83 m/s^2. Using the same equation, we can calculate the time for 100 oscillations: T = 2*pi*sqrt(1.000/9.83) = 2*pi*0.321 = 2.01 seconds. Therefore, 100 oscillations would take 2.01*100 = 201 seconds at the north pole.

(c) At the top of a high mountain peak near the equator, the acceleration due to gravity may be slightly different due to the change in altitude. Let's say we measure the time for 100 oscillations to be 201 seconds. Using the same equation, we can solve for g: g = 4*pi^2*L/T^2 = 4*pi^2*1.000/(201/100)^2 = 9.71 m/s^2. Therefore, the acceleration due to gravity on the mountain top is 9.71 m/s^2.

## What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which a system oscillates back and forth around an equilibrium point, with a restoring force that is directly proportional to the displacement from the equilibrium point.

## What is a pendulum?

A pendulum is a weight suspended from a fixed point that swings back and forth under the influence of gravity. It is a common example of simple harmonic motion.

## What factors affect the period of a pendulum?

The period of a pendulum is affected by its length, the acceleration due to gravity, and the angle at which it is released.

## How does the length of a pendulum affect its period?

The period of a pendulum is directly proportional to its length. This means that a longer pendulum will have a longer period and a shorter pendulum will have a shorter period.

## What is the relationship between the period and frequency of a pendulum?

The period of a pendulum is the time it takes for one full oscillation, while the frequency is the number of oscillations per unit of time. The period and frequency are inversely related, meaning that as one increases, the other decreases.

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