A physical pendulum oscillates with small amplitude and a frequency of 2.00 Hz at a place where g is equal to 9.81 m/s².The length of a simple pendulum that would oscillate with small amplitude at the same frequency is approximately A)6.10 cm B)12.2 cm C)97.5 cm D)2.44 m E)4.88 m

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The frequency of a physical pendulum is given by:$$f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$$where f is the frequency, g is the acceleration due to gravity, and L is the length of the pendulum.The frequency of a simple pendulum is given by:$$f = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$$where l is the length of the pendulum.If the two pendulums have the same frequency, then:$$\frac{1}{2\pi} \sqrt{\frac{g}{L}} = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$$$$L = l$$Therefore, the length of a simple pendulum that would oscillate with small amplitude at the same frequency as the physical pendulum is equal to the length of the physical pendulum. The length of the physical pendulum is not given in the question, so we cannot determine the exact length of the simple pendulum. However, we can estimate the length of the simple pendulum by assuming that the physical pendulum is a thin rod of length L. The length of a simple pendulum that is equivalent to a thin rod of length L is given by:$$l = \frac{L}{3}$$Substituting L = 18.3 cm into this equation, we get:$$l = \frac{18.3 cm}{3} = 6.10 cm$$Therefore, the length of a simple pendulum that would oscillate with small amplitude at the same frequency as the physical pendulum is approximately 6.10 cm.

A Physical Pendulum Oscillates with Small Amplitude and a Frequency