A mass m1 hangs from a spring k N/m and is in a static equilibrium. A second mas

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 cm

Explanation / 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

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