A child releases himself from a tree branch 50 cm above a platform onto the plat

Question

A child releases himself from a tree branch 50 cm above a platform onto the platform. THe platform is mounted onto a spring. Together, they come to rest 10 cm below the initila position of the platform. a.) Assume the masses ofthe child and the platform are 10 kg and 2 kg, respectively. What is the spring constant ofthe spring? b.) Assume that the platform-spring system is designed such that it would bring a 10 kg child to rest in minimum time without overshooting. What is the damping factor? c.) Find the equation of motion for the platform, choosing upward as the positive direction, the final rest position as the origin of the x-coordinate, and the child's impact instant as t=0. Additionally, treat the child's impact to the platform as a perfect inelastic collision between the child and the platform.

Explanation / Answer

given distance of the child form the platform in vertical direction, h = 50 cm = 0.5 m

so when the person drops, total compression of the spring = x = 0.1 m

now from conservation of energy

Mg(h + x) + mgx = 0.5kx^2

where m is mas so of person, M = 10 kg

m = 2 kg ( mass of platform)

k = spring constant of the spring

hence 10*g(0.6) + 2*g*(0.1) = 0.5*k*0.1^2

a. k = 12164.4 N/m

b. for the child to come to rest in minimum time, the system has to be critically damped

so c^2 = 4(m+ M)k

c = sqroot((m + M)k)*2 = 764.1227 Ns / m

d. the equation of motion for a damped harmonic osscilator is given by

mx" + cx' + kx = 0

where x = A*e^(lambda*t) + B

at t = 0, x = 0

t = 0, x' = Vo = sqroot(2gh) = 3.132 m/s

so, x = A + B = 0

A*lambda = 3.132

so, x = A(e^(lambda*t) - 1_

where A*lambda = 3.132 and lambda is a solution to lambda^2*m + lambda*x + k = 0

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