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Figure 16.46 This child's toy relies on springs to keep infants entertained. (credit: By Humboldthead, Flickr) The device pictured in Figure 16.46 entertains infants while keeping them from wandering. The child bounces in a harness suspended from a door frame by a spring constant. (a) If the spring stretches 0.250 $\mathrm{m}$ while supporting an 8.0 -kg child, what is its spring constant? (b) What is the time for one complete bounce of this child? (c) What is the child's maximum velocity if the amplitude of her bounce is 0.200 $\mathrm{m} ?$

$\begin{aligned} K=& 313.6 \mathrm{N} / \mathrm{m} \\ & T=1 \mathrm{s} \\ V &=2.36 \mathrm{m} / \mathrm{s} \end{aligned}$

Physics 101 Mechanics

Physics 103

Chapter 16

Oscillatory Motion and Waves

Periodic Motion

Wave Optics

Rutgers, The State University of New Jersey

Hope College

University of Sheffield

McMaster University

Lectures

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you know that. Here, um, the string const The weight of the baby is going to be equal to the force of the spring. This would be equaling for party. Angie. The weight is equaling negative K x, and so we can find the spring constant k. This would be negative. Maso negative 8.0 kilograms multiplied by 9.80 meters per second squared, divided by X negative 0.250 meters. And this is equaling 313.6 newtons per meter. This would be our answer. Four part, eh, four part B. We want to find the period. This is equaling two pi times the square root of the mass divided by K the spring constant. This would be to pie multiplied by the square root of 8.0 kilograms, divided by 313.6 newtons per meter. And this is equaling 1.0 seconds. And so this would be equal to the period. And then finally, for part C, we confined their maximum velocity. We can relate the potential energy of a spring, see, be equal to the maximum kinetic energy and so weaken this would be squared. We can cancel out the factor of 1/2 and weak and then say that the maximum velocity would be equaling, uh, ex multiplied by the square root of the spring constant divided by the mass M. And so we find that four part see, the maximum velocity is equaling a square root of 313.6 newtons per meter divided by 8.0 kilograms. And then this would be multiplied by 0.200 meters, the maximum displacement from equilibrium. And we find that the maximum velocity is equaling 1.25 meters per second. This would be our final answer for part C. That is the end of the solution. Thank you for watching.

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