Answered

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58) A simple pendulum of length 2.3 m makes 5.0 complete swings in 37 s. What is the acceleration of gravity at the location of the pendulum?

Sagot :

Take into account that the period of a pendulum is given by the following formula:

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

where,

l: length = 2.3 m

g: acceleration of gravity = ?

T: period

The period T is:

[tex]T=\frac{37s}{5.0}=7.4s[/tex]

If you solve for g the equation of the period, you obtain:

[tex]\begin{gathered} T^2=4\pi^2\frac{l}{g} \\ g=\frac{4\pi^2l}{T^2} \\ g=\frac{4\pi^2(2.3m)}{(7.4s)^2}\approx1.66\frac{m}{s^2} \end{gathered}[/tex]

Hence, the acceleration of gravity at the location of the pendulum is approximately 1.66m/s^2

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