Answer:
7.50 m/s^2
Explanation:
The period of a pendulum is given by:
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex] (1)
where
L = 0.600 m is the length of the pendulum
g = ? is the acceleration due to gravity
In this problem, we can find the period T. In fact, the frequency is equal to the number of oscillations per second, so:
[tex]f=\frac{N}{t}=\frac{20}{35.5 s}=0.563 Hz[/tex]
And the period is the reciprocal of the frequency:
[tex]T=\frac{1}{f}=\frac{1}{0.563 Hz}=1.776 s[/tex]
And by using this into eq.(1), we can find the value of g:
[tex]g=\frac{4 \pi^2 L}{T^2}=\frac{4 \pi^2 (0.600 m)}{(1.776 s)^2}=7.50 m/s^2[/tex]
