A building in San Francisco has light fixtures consisting of small 2.35-kg bulbs with shades hanging from the ceiling at the end of light, thin cords 1.50 m long. If a minor earthquake occurs, how many swings per second will these fixtures make?

Video transcript

Hey, everyone in this problem, a grandfather clock with a 1 m long pendulum hangs on a clock tower. The mass of the bulb on the pendulum is 0.5 kg. An unlucky driver hits the tower vibrating it. If the pendulum was initially at rest in the equilibrium position, how many cycles per second will the pendulum make? Now when we hear cycles per second? Hm Well, to recall that this is going to be equal to the frequency F, by definition, the frequency is the number of cycles per second. So we're looking for the frequency gap. Now, we have some information about the length of the clock. Let's recall that we can relate the frequency to the length through the follow two pi times the frequency F is equal to the square root of G divided by L the length of the pendulum. All right. So we want to find the frequency EF G is gonna be the gravitational acceleration on earth, which we know and L is the length of the pendulum, which we're told. So we can find F OK. So F is going to be equal to 1/2 pi times the square root of G over L. This is gonna be 1/2 pi times the square root of 9.8 m per second squared divided by the length L which is 1 m. And this gives us, gives us a frequency F of 0.49 eight Hertz. OK. So if we're looking at the number of cycles per second, OK. That's the same as the unit of Hertz, same as a frequency. Ok. So we found that we're gonna be making 0.498 cycles per second. Ok. So we have answer B thanks everyone for watching. I hope this video helped see you in the next one.