After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 49.0 cm. The explorer finds that the pendulum completes 109 full swing cycles in a time of 144 s.What is the value of the acceleration of gravity on this planet?

Respuesta :

The frequency refers to the number of cycles per time lapse, while the period is the inverse of the frequency, therefore we would have this expression,

[tex]f= \frac{N}{t}[/tex]

Here,

N = Number of Cycles

t = Time

While at the same time Period is the inverse of the frequency, then

[tex]T = \frac{1}{f}[/tex]

Replacing,

[tex]T = \frac{1}{N/t}[/tex]

[tex]T = \frac{t}{N}[/tex]

Replacing we have that

[tex]T = \frac{144s}{109}[/tex]

[tex]T = 1.321s[/tex]

Gravity is one of the components that define the Period of a pendulum as well as its length, mathematically this expression is given as,

[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]

[tex]g = \frac{4\pi^2l}{T^2}[/tex]

Replacing,

[tex]g = \frac{4\pi^2 (49*10^{-2}m)}{(1.321s^2)^2}[/tex]

[tex]g = 11.0853m/s^2[/tex]

Therefore the acceleration due to gravity of the planet is [tex]11.08m/s^2[/tex]