The device pictured in the figure entertains infants while keeping them from wan
ID: 1402760 • Letter: T
Question
The device pictured in the figure entertains infants while keeping them from wandering. The child bounces in a harness, with a spring constant k, and is suspended off of the ground.Randomized Variablesx1 = 0.225 m
A = 0.215 m
m = 8.35 kg (a) If the spring stretches x1 = 0.225 m from equilibrium while supporting an 8.35-kg child, what is its spring constant, in newtons per meter?
(b) What is the time, in seconds, for one complete bounce of this child about the point x1? (c) What is the child’s maximum velocity, in meters per second, if the amplitude of her bounce, relative to x1, is 0.215 m?
Explanation / Answer
Gievn that,
max strech = x1 = 0.225 m ; mass of the child = m = 8.35 kg ; Amplitude = 0.215 m
We know that, from hook's law, F = - Kx this is getting balanced with the weigth. So
K x1 = mg
K = mg/x1 = 8.35 x 9.8 / 0.225 = 363.7 N/m
Hence, K = 363.7 N/m
(b)Let "T" be the time and it will be given by:
T = 2 pi sqrt(m/K) = 2 x 3.14 x sqrt (8.35/363.7) = 0.952 seconds
Hence, the time = T = 0.952 sec.
(c)Let v be the velocity.
We know that the potential energy stored in the spring is
The potential energy of the spring with amplitude A will be
PE(amp A) = 1/2 K A2 = 0.5 x 363.7 x (0.215)2 = 8.41 J
This should be equal to the KE of child, so
1/2 m v2 = 8.41 => v = sqrt(2 x 8.41 / 8.45 ) = 1.41 m/s
Hence, the velocity = v = 1.41 m/s.