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A uniform meter stick is freely pivoted about the 0.20-m mark. If it is allowed to swing in a vertical plane with a small amplitude and friction, what is the frequency of its oscillations? A) 0.66 Hz B) 0.92 Hz C) 0.55 Hz D) 1.1 Hz E) 1.3 Hz

Respuesta :

Option C is the correct answer.

Explanation:

Length of pendulum, l = 1 - 0.2 = 0.8 m

We have equation for period of pendulum

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

Substituting length value

                         [tex]T=2\pi \sqrt{\frac{l}{g}}\\\\T=2\pi \sqrt{\frac{0.8}{9.81}}\\\\T=1.79s[/tex]

Frequency is the reciprocal of time period,

                [tex]f=\frac{1}{T}=\frac{1}{1.79}=0.55Hz[/tex]

Option C is the correct answer.

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