Question

The frequency of a person's pendulum is 0.3204 Hz when at a location where g is known to be exactly 9.800 m/s^2.
A) What is g in the location where the same pendulum's frequency is 0.3196 Hz?
Express your answer to four significant figures and include the appropriate units.
B) Assume this g is in the location strictly above the initial location (with g = 9.800 m/s^2). What is the height of this location?
Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

A) g = 9.751 m/s², B)       h = 2.573 10⁴ m

Explanation:

The angular velocity of a pendulum is

           w = √ g / L

Angular velocity and frequency are related.

          w = 2π f

          f = 1 / 2π √ g / L

A) with the initial data we can look for the pendulum length

        L = 1 /4π²   g / f²

        L = 1 /4π²   9,800 / 0.3204²

        L = 2.4181 m

The length of the pendulum does not change, let's look for the value of g for the new location

        g = 4π² f²  L

        g = 4π² 0.3196²  2.4181

        g = 9.75096 m / s²

         g = 9.751 m/s²

B) The value of the acceleration of gravity can be found with the law of universal gravitation

           F = G m M / $R_{e}$²

And Newton's second law

           W = m g

           W = F

         G m M / $R_{e}$² = mg

         g = G M / $R_{e}$²

         $R_{e}$² = G M / g

Let's calculate

          $R_{e}$² = 6.67 10⁻¹¹ 5.98 10²⁴ /9.75096

          R = √ 4.0905 10¹³ = √ 40.9053 10¹²

          R = 6.395726 10⁶ m

The height above sea level is

         h = R - [tex]R_{e}[/tex

         h = (6.395726 -6.37) 10⁶

         h = 0.0257256 106

         h = 2.573 10⁴ m