Question

A 0.500-kg mass suspended from a spring oscillates with a catamenia of 1.50 s. How much mass must be added to the object to change the period to 2.00 due south?

Solution Video

OpenStax College Physics Solution, Chapter xvi, Problem 15 (Issues & Exercises) (2:09)

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Video Transcript

This is College Physics Answers with Shaun Dychko. The menses of a unproblematic harmonic oscillator is two Pi times the square root of mass divided by the spring abiding. Now nosotros have the flow in the first instance is ii Pi foursquare root m one over thousand. And and so in the second instance after some mass is added that's delta m. Information technology'south going to be two Pi times foursquare root of the original mass plus the amount that gets added divided by chiliad. And our job here is to solve for delta m. So we'll take menstruation two which we know is 2 seconds, divided by period one which is one and a half seconds. And it's going to equal two Pi times square root m1 plus delta m over g. And and so instead of dividing past this fraction, I'm going to multiply information technology by its reciprocal. And then when dividing by T1, we're going to write instead multiplied by one over ii Pi times square root g over m1. That's the reciprocal of this. I'm writing it that way makes information technology easy to come across that the two Pi is cancel and the square root k is cancel. And this gives T2 over T1 is square root of m1 plus delta m over m1. Then we square both sides to get rid of the square root sign and we accept T2 over T1 squared equals m1 plus delta thou over m1. So we'll multiply both sides by m1 and so switch the sides around so the unknown is on the left. And we have m1 plus delta m equals m1 times T2 over T1 squared. And so we decrease m1 from both sides to solve for delta m. So delta grand which is the amount of mass that must exist added to increment the period to ii seconds is m1 times T2 over T1 squared minus m1. Then we could gene the m1 out, although we don't really have to merely information technology's simply information technology looks a lilliputian neater that way. And then, substituting in numbers, we take delta thou that is 0.v kilograms times 2 seconds which is the period after adding some mass divided by one and a one-half seconds. And that is squared and then minus one. And we become 0.38 kilograms every bit the corporeality of mass that must be added.

Solutions for issues in chapter 16