A block attached to a spring with unknown spring constant oscillates with a period of 2.0 s. What is the period if d. The spring constant is doubled? Parts a to d are independent questions, each referring to the initial situation.

Video transcript

Hey, everyone. So this problem is dealing with simple harmonic motion. Let's see what it's asking us. A 300 g mass is attached to a spring that experiences oscillations with a period of 1.5 seconds. Determine the period of oscillations. If the spring constant is made four times its original value. Our multiple choice answers here are a six seconds. B 0.75 seconds, C three seconds or D 0.38 seconds. OK. So the key to this problem is to recall two equations. One for the period, which is given as two T equals two pi divided by Omega. And the second equation for the spring constant K equals mass multiplied by Omega squared omega is that angular frequency. And so we can kind of merge these equations together. We can solve the second equation for Omega and then plug that in two hour period because they're asking for the period if that friend constant is four times higher. So if we isolate omega, we get the square root of K divided by M like that and you get T or period is equal to two pi divided by the square root of K divided by M. And so in our first scenario or T one, that's, that's just all that it is. We'll look at our second scenario T two and that equals two pi died by the square root of four K divided by that, you can split out this two pi divided by the square root of K divided by M. And then we are left with just the S word of four in the denominator. And so this whole piece is equal to T one which we know from the problem is 1.5 seconds. So let me plug that in. We get 1.5 seconds multiplied by one over squared of four which is two or just one half instead of 1.5 seconds multiplied by a half is 0.75 seconds. And that is the answer to this problem. And so that aligns with answer choice B so that's all we have for this one. We'll see you in the next video. Mhm.