Answer :
The period of a simple pendulum is given by:
[tex]T=2 \pi \sqrt{ \frac{L}{g} } [/tex]
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
[tex]g= \frac{4 \pi^2}{T^2}L [/tex] (1)
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
[tex]f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz [/tex]
And its period is the reciprocal of its frequency:
[tex]T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s [/tex]
So now we can use eq.(1) to find the gravitational acceleration of the planet:
[tex]g= \frac{4 \pi^2}{T^2}L = \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2[/tex]
[tex]T=2 \pi \sqrt{ \frac{L}{g} } [/tex]
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
[tex]g= \frac{4 \pi^2}{T^2}L [/tex] (1)
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
[tex]f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz [/tex]
And its period is the reciprocal of its frequency:
[tex]T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s [/tex]
So now we can use eq.(1) to find the gravitational acceleration of the planet:
[tex]g= \frac{4 \pi^2}{T^2}L = \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2[/tex]