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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 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 m?

Answer :

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The child’s maximum velocity if the amplitude of her bounce is 0.200 m is,

  • spring constant is 32 N/m.
  • the time for one complete bounce of this child is  1.003 sec.
  • child’s maximum velocity we get  .2006 m/s

What is the spring constant ?

  • The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies. The stiffer the spring is and the harder it is to stretch, the larger the spring constant.

weight of the infants = 8 Kg

mass =8/9.8 =0.82 Kg

the spring stretches  = 0.250 m

W = kx.

k = w/x = 8/0.25 =32 N/m

spring constant is 32 N/m

A mass on a spring will oscillate with a period if it is coupled to a spring with spring constant k. (T). describes this. We can calculate the period and, consequently, the frequency by measuring the length of one complete oscillation.

T = 2π√(m/k) =  2π√0.82/32 = 2π * 0.160 = 1.003 sec.

max velocity

v(t)=dxdt

dx= 0.200

dt = 1.003

v(t) = 0.200 * 1.003 =.2006 m/s

To learn more about spring constant,  refer,

brainly.com/question/23885190

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