a block of mass m= 3kg is suspended by a spring of constant k= 2700 N/m
Assume that +y direction is upward. if the mass is moving downward through its equilibruim position with speed v=6m/s at time t=0, the position (in m) of the mass as a function of time (in s ) with respect to its equilibrium position will be given by?
1- .2 cos (30t+pi/2)
2- .2cos(30t-pi)
3- 6 sin(30t-pi)
4- 180 cos(30t+pi/2)
5- .2 sin(900t+pi)
choose the right answer and explain
a block of mass m= 3kg is suspended by a spring of constant k= 2700 N/m...
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5 points) A spring is suspended vertically from a fixed support.
The spring has spring constant k=28 N m−1k=28 N m−1. An object of
mass m=14 kgm=14 kg is attached to the bottom of the spring. The
subject is subject to damping with damping constant β N m−1 sβ N
m−1 s. Let y(t)y(t) be the displacement in metres at the end of the
spring below its equilibrium position, at time tt seconds.
(5 poins) A spring is suspended vertically...
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(5 points) A spring is suspended vertically from a fixed
support. The spring has spring constant k=28 N m−1k=28 N m−1. An
object of mass m=14 kgm=14 kg is attached to the bottom of the
spring. The subject is subject to damping with damping constant β N
m−1 sβ N m−1 s. Let y(t)y(t) be the displacement in metres at the
end of the spring below its equilibrium position, at time tt
seconds.
(a) Give a value of ββ which...
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