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A massless rod of length L is hinged with a frictionless hinge at one end. At the end opposite to the hinge there is a very small mass m attached to the rod. The rod is released from rest in the horizontal position. What is the frequency f when the rod passes through the vertical position?1. 2πSQRT(g/L)2.2g/L3.(1/2π)SQRT(2g/L)4. (1/2π)SQRT(g/L)5. 2πSQRT(2g/L)6. SQRT(2g/L)7.g/L8. SQRT(g/L)

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whitneytr12

Answer:

4. [tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}[/tex]

Explanation:

We have in this case a simple pendulum, of length L and mass m. As we know, the period of this pendulum is:

[tex]T\sim 2\pi \sqrt{\frac{L}{g}}[/tex]

but it is true only in small oscillation angles, it meas less than 1 rad.

The frequency is the inverse of the period, then:

[tex]f=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{g}{L}}[/tex]

The correct answer is 4.

I hope it helps you!

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