A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, with the free end of the spring at the pivot. What rotation frequency, in rpm, will cause the spring's length to stretch by 15%?

Question

Accepted Answer

Hey, everyone. So this problem is dealing with simple harmonic motion. Let's see what it's asking us. We have a spring mass system on a horizontal disc initially at rest. The spring's free end is attached to the disc axle. The math undergoes periodic motion with a period of 0. seconds when the disc is spun at a certain frequency where the mass is not oscillating, the spring extends and its length is increased by 25%. We're asked to calculate the dick the discs rotational frequency in rotations per minute or R PM. So our multiple choice answers here are a 15 R PM B 24 R P MC, 38 R PM or D 46 R PM. So the first thing we can do for this problem is apply mutant second law in the radial direction. And so that looks like the, some of the forces in the radial direction is equal to the mass times the radial acceleration. And so the only force acting in the uh real direction is that spring force. So we have K multiplied by uh delta R or a change in distance and that equals mass multiplied by. We can recall that in simple harmonic motion, our acceleration can be rewritten as R multiplied by omega squared. And so isolating Omega in this equation, we get the square root of K divided by M multiplied by delta R divided by R and splitting it up. Like that is important because the problem is talking about these lengths, these, this what the amount that the spring extends, but it doesn't um give you actual values, it just talks about them relatively. So when we look at what delta are divided by our equals, so delta R is gonna be R minus R not. And so then divided by R, I know that that is going this initial R stretched R S B 1.25 R minus R and then just divided by R gives us 0.25 R and the A canceled. So we had just 0.0 point 25. And so we've figured out how to solve for this half of the equation. And now we need to look at this a divided by M term. We can recall that for uh simple harmonic motion or period is given by the equation T equals two pi multiplied by the square root of M divided by K. And so getting it into that square root of K divided by M that equals to pi divided by T. And so we now have, when we insert dot N two, um our Omega equation or our radial frequency equation. We now have everything we need to solve for our radio frequency. So we'll have two pi divided by T multiplied by delta R divided by R and squared up all of that you plug in those known values. So, too high tea was given to us in the problem at 0.65 seconds. And then we've already determined that delta R divided by R is 0.25. And so our sorry have a um square root incorrect there. So the square root is actually just over that um delta R O. So I apologize, let's clean that up really quickly there. OK. So let me plug that correct equation, those correct values into our calculator. We get 4.83 radians per second. Now, the last step here is recalling that they asked us to give our answer in revolutions per minute. And so we can recall that one revolution is equal to two pi radians and then you take it from seconds to minutes to 60 seconds per minute. And our final answer for our frequency is 46 R PM. And that aligns with answer choice D. So that's all we have for this one. We'll see you in the next video.

A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, with the free end of the spring at the pivot. What rotation frequency, in rpm, will cause the spring's length to stretch by 15%?

Verified Solution

Key Concepts

Simple Harmonic Motion (SHM)

Centripetal Force

Spring Constant and Hooke's Law