Apply the impulse-momentum relation and the work-energy theorem to calculate the maximum value of t if the cake is not to end up
1 answer:
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Answer:
(a) The period is 4s
(b) The angular frequency is pi/2 radians
(c) The amplitude is 0.37cm.
(d) The displacement at time is (0.37 cm) cos((pi/2)*t)
(e) The Velocity at time t is v = (0.58 cm)(sin((pi/2)*t)
(f) The maximum speed is
(g) The maximum acceleration is 0.91 cm/s^2
Explanation:
We have a particle which oscillates with frequency of f = 0.25 Hz about the point x = 0.At t = 0, the displacement of the particle is = 0.37 cm and its velocity is zero.
(a) The period of the oscillations is,
so
T = 1/(0.25 Hz)
T = 4.0s
(b) The angular frequency is,
f = = 2()(0.25 Hz) =
radians
(c) Since
The amplitude is the maximum displacement that the particle makes from the equilibrium point, or when the speed of the particle is zero,
that is
= 0.37 cm
(d) The displacement as a function of t is given be,
x = cos(ωt+Φ)
as x = t = 0, we get cos(Ф) = 1 = 0
so this equation becomes
x= (0.37 cm) cos((pi/2)*t)
(e) Now we need to find the speed of the particle as a function of t
the speed is the derivative of the displacement that is
= -(0.37)(pi/2)(sin((pi/2)*t)
so the velocity at time t is
v = (0.58 cm)(sin((pi/2)*t)
(f) Since
sin(ωt+Ф)
then from part (e) we get
(g)
The amplitude of the maximum acceleration is
ω^2
= (0.37 cm) (pi/2) = 0.91 cm/s^2
this is the maximum acceleration