Estimate the stiffness of the spring in a child’s pogo stick if the child has a mass of 32 kg and bounces once every 2.0 seconds.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The frequency of oscillation on this child in the pogo stick is 1 over 2π times the square root of the spring constant divided by the child-pogo stick combined mass. We'll assume that the pogo stick has negligible mass and so m will just be the mass of the child. Multiply both sides by 2π and switch the signs around and we get square root k over m as 2π f and then square both sides and get k over m in the left equals 4π squared f squared on the right. And then multiply both sides by m. And you get k is 4π squared f squared times m. And the frequency is one cycle for every 2 seconds. And that gives you cycles per second or hertz in other words. I could have used the period formula, maybe that would've been a little bit better but in any case it doesn't really matter. You could also think of this f as being a reciprocal of period. And so that's basically what we have here. We have one cycle divided by the time for one cycle which is 2 seconds. And times 32 kilograms. And all that works out to about 320 newtons per meter for the spring constant.