1. A particle undergoes damped harmonic motion. The spring constant is 147 N/m;

Question

1. A particle undergoes damped harmonic motion. The spring constant is 147 N/m; the damping constant is 0.005 kg·m/s, and the mass is 0.06 kg. If the particle starts at its maximum displacement, x = 1.7 m, at time t = 0, what is the amplitude of the motion at t = 7 s??

2. A hoop (I = MR2) of mass 3 kg and radius 1 m is rolling at a center-of-mass speed of 10 m/s. An external force does 810 J of work on the hoop. What is the new speed of the center of mass of the hoop (in m/s)? Round your answer to the nearest whole number.

3. A uniform thin rod is hung vertically from one end and set into small amplitude oscillation. If the rod has a length of 2.7 m, this rod will have the same period as a simple pendulum of length ____cm.

4. An Russian spy satellite nears the end of its life due to air resistance. Before it meets its demise: (please explain too)

a.it moves slower as the orbit lowers
b.it moves faster in the same orbit but with an increasing period
c.it moves slower in the same orbit but with a decreasing period
d.it slowly spirals away from Earth
e.It moves faster as the orbit lowers

5. A particle is located on the x axis 4.5 m from the origin. A force of 31 N, directed 30° above the x axis in the x-y plane, acts on the particle. What is the torque about the origin on the particle? Round your answer to the nearest whole number. Use a minus sign to indicate a negative direction and no sign to indicate a positive direction.

Explanation / Answer

1.

for the damped harmonic motion the Amplitude equation is A(t) = Ao e -bt/2m

b = damping constant

m = mass

A(7) = 1.7 * e - ( ( 0.005 * 7 ) / 2 * 0.06 )

   = 1.7 * e -0.291

   = 1.27 m is answer

please ask other questions as a saperate question

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