Asked by gill
A baby bounces up and down in her crib. Her mass is 13.5 kg, and the crib mattress can be modeled as a light spring with force constant 682 N/m.
If she were to use the mattress as a trampoline—losing contact with it for part of each cycle—what minimum amplitude of oscillation does she require?
If she were to use the mattress as a trampoline—losing contact with it for part of each cycle—what minimum amplitude of oscillation does she require?
Answers
Answered by
Damon
w = 2 pi f = sqrt(k/m)
x = A sin wt
v = A w cos wt
a = -A w^2 sin wt
max a = A w^2 = g
so
g = A w^2
A = g m/k
x = A sin wt
v = A w cos wt
a = -A w^2 sin wt
max a = A w^2 = g
so
g = A w^2
A = g m/k
Answered by
Anonymous
to damon. that doesnt help. be more detailed.
Answered by
ANON
Damon is Stating that-
F=1/T T=2*pi/w so T= 2pi/sqrt(k/m)
using our values T=2pi/sqrt(682/13.50)=.884004
Therefore F=1/.884004 -->=1.13121
For the second part acceleration=Aw^2 which must equal the acc. of gravity(g)
g=Aw^2 w=sqrt(k/m) so w=7.107638
plugging into our equation the equation g=Aw^2: 9.8=A*(7.107638^2)
9.8=50.518518*A
9.8/50.518518=A, so A=.193988 meters
which is about 19.39 cm
F=1/T T=2*pi/w so T= 2pi/sqrt(k/m)
using our values T=2pi/sqrt(682/13.50)=.884004
Therefore F=1/.884004 -->=1.13121
For the second part acceleration=Aw^2 which must equal the acc. of gravity(g)
g=Aw^2 w=sqrt(k/m) so w=7.107638
plugging into our equation the equation g=Aw^2: 9.8=A*(7.107638^2)
9.8=50.518518*A
9.8/50.518518=A, so A=.193988 meters
which is about 19.39 cm
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