Complete Question
The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2
Answer:
The acceleration due to gravity is [tex]g = 167.2 \ m/s^2[/tex]
Explanation:
From the question we are told that
The length of the simple pendulum is [tex]L = 1.081.08 \ m[/tex]
The number of cycles is [tex]N = 101[/tex]
The time take is [tex]t = 2.00 *10^{2 \ }s[/tex]
Generally the period of this oscillation is mathematically evaluated as
[tex]T = \frac{N}{t }[/tex]
substituting values
[tex]T = \frac{101}{2.0*10^2 }[/tex]
[tex]T = 0.505 \ s[/tex]
The period of this oscillation is mathematically represented as
[tex]T = 2 \pi \sqrt{\frac{l}{g} }[/tex]
making g the subject of the formula we have
[tex]g = \frac{L}{[\frac{T}{2 \pi } ]^2 }[/tex]
[tex]g = \frac{4 \pi ^2 L }{T^2 }[/tex]
Substituting values
[tex]g = \frac{4 * 3.142 ^2 * 1.08 }{505.505^2 }[/tex]
[tex]g = \frac{4 * 3.142 ^2 * 1.08 }{0.505^2 }[/tex]
[tex]g = 167.2 \ m/s^2[/tex]
