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A 0.600 m long pendulum is used to determine the acceleration due to gravity on a distant planet. If 20 complete oscillation are completed and 35.5 seconds, the acceleration is

a) 7.52 m/s squared

b) 9.81 m/s squared

c) 30.5 m/s squared

d) 42.0 m/s squared

Respuesta :

Answer :

Explanation :

It is given that,

length of pendulum, l = 0.6 m

Number of oscillation, n = 20

time, t = 35.5 seconds

so, Time period, [tex]T=\dfrac{t}{n}[/tex]

[tex]T=\dfrac{35.5s}{20}[/tex]

we know that the time period of pendulum is given by

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

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

[tex]g=\dfrac{4\times(3.14)^2\times 0.6 \times (20)^2}{(35.5s)^2}[/tex]

[tex]g=7.51\ m/s^2[/tex]

or

[tex]g=7.52\ m/s^2[/tex]

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