A mass m1 hangs from a spring k N/m and is in a static equilibrium. A second mas
ID: 1854694 • Letter: A
Question
A mass m1 hangs from a spring k N/m and is in a static equilibrium. A second mass m2 drops through a height h and sticks to m1 without rebound, as shown in the accompanying figure. Determine the subsequent motion. What will be the maximum amplitude of this motion when the masses, stiffness and the height of drop have the following values? Case 1: mj = 10.5 kg m2 = 20.4 kg k = 25kN/m h = 35 cm Case 2: mi = 20.4 kg m2 = 10.5 kg k = 25kN/m h = 35 cm Case 3: mi = 20.4 kg m2 = 10.5 kg k = 27.5kN/m h = 35 cm Case 4: mi = 20.4 kg m2 = 10.5 kg k = 25kN/m h = 70 cmExplanation / Answer
as m2 droped will contribute accelration a=g always indipendent of h
thus
kx=m1g+m2g+m2a
kx=g(m1+2m2)
x=g(m1+2m2)/k
case 1:
m1=10.5kg
m2=20.4kg
k=25kN/m
h=35cm=0.35m
u=0
since
x=0.02m
max amp=x+h=0.37m
case 2
m1=20.4
m2=10.5
k=25kN/m
h=35cm=.35m
x=0.0162m
max amp=x+h=0.3662m
case3:
m1=20.4kg
m2=10.5kg
k=27.5kN/m
h=35cm=0.35m
x=0.1475m
max amp=x+h=.36475m
case4:
m1=20.4kg
m2=10.5kg
k=25kN/m
h=70cm=0.7m
x=0.0162m
max amp=x+y=0.7162m