After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. The explorer finds that the pendulum completes 93.0 full swing cycles in a time of 129 s.

What is the value of the acceleration of gravity on this planet?

Answer :

Answer:

The value of the acceleration of gravity on this planet is [tex]11.4\ m/s^2[/tex].  

Explanation:

Given that,

Length of the simple pendulum, l = 55 cm = 0.55 m

The explorer finds that the pendulum completes 93.0 full swing cycles in a time of 129 s. The time period is given by :

[tex]T=\dfrac{129}{93}=1.38\ s[/tex]

We need to find the value of the acceleration of gravity on this planet. The time period in case of simple pendulum is given by :

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

[tex]g=\dfrac{4\pi ^2\times 0.55}{(1.38)^2}\\\\g=11.4\ m/s^2[/tex]

So, the value of the acceleration of gravity on this planet is [tex]11.4\ m/s^2[/tex]. Hence, this is the required solution.

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