The 15 g head of a bobble-head doll oscillates in SHM at a frequency of 4.0 Hz. b. The amplitude of the head's oscillations decreases to 0.5 cm in 4.0 s. What is the head's damping constant?

Video transcript

Hey, everyone. So this problem is dealing with simple harmonic motion. Let's see what they're asking us. We have a mass of 25 g attached to a spring that undergoes simple harmonic motion with a frequency of two Hertz and an amplitude of 10 centimeters. If the amplitude decreases to six centimeters after five seconds due to damping. What is the damping constant of the system? Our multiple choice answers here are a 0. kg per second. B 0.5 kg per second. C 0.6 kg per second or D 0.7 kg per second. So the key to solving this problem is recalling that the damped oscillations, the position of those damp ations is given by the equation X of T equals a knot multiplied by E to the negative B multiplied by T divided by two M. We can soul this problem to isolate B which is our damping constant. So our X of T, so our position at five seconds is going to be six centimeters oops. So six centimeters is equal to. There we go. Our A not our original position of centimeters multiplied by E to the negative B time five seconds divided by two times our mass, which was given us 25 g. We're going to write that in standard units as 0.25 kg. The reason why I didn't bother to put the six and the centimeters in standard units is because we're going to divide those units cancel. Um If you wanted to keep everything in standard units, obviously, you could have done that too. And so when we divide each side by 10 centimeters, we get 100.6 is equal to E to the negative 100 B. So then we need to use our L N rules. So L N of 0.6 divided by negative is equal to B and that equals 0.5 kilograms per second. And so that is the final answer and it aligns with answer choice. A sorry answer. Choice. B All right. That's all we have for this one. We'll see you in the next video.