Question

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)

Answer 1

4.
`f=(1)/(2\pi)\sqrt{(g)/(L)}`

Step-by-step explanation:

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

`T\sim 2\pi \sqrt{(L)/(g)}`

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

The frequency is the inverse of the period, then:

`f=(1)/(T)=(1)/(2\pi)\sqrt{(g)/(L)}`

The correct answer is 4.

I hope it helps you!