Answered: 3. A 0.55 kg vibrates according to the… | bartleby

Related questions

Question

3. A 0.55 kg vibrates according to the equation x = 1.5 cos (12.4t). a) determine the equations of the velocity and
accelerations as a function of time. b) what is the velocity, acceleration and restoring force at x = 0.5m?

Expert Solution

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

A 500g block connected to a spring which the force constant is 5 N/m is free to oscillate on a frictionless, horizontal surface. The block is displaced 6cm from equilibrium and released from rest. Calculate the following: a.Period ofthe motion in seconds b.Maximum speed in m/s. c.Maximum acceleration in m/s2 d.Express the position, velocity, and acceleration as functions of time.
A 0.500 kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the path at a constant frequency of 2.20 rev/sec. At the moment the stone is overhead, the stone is released. The velocity of the stone when it leaves the circular path is Multiple Choice 10.4 m/s horizontal. 10.4 m/s vertical. 22.0 m/s horizontal. 22.0 m/s vertical. 31.4 m/s horizontal.
A 5.50 meter spring is pulled with a force of 58N and stretches by 0.25m, to a total length of 5.75m. When released, it vibrates at 13 rad/s. What is the spring constant of the spring?Do not actually solve the problem numerically or algebraically, just pick the one equation and define the relevant knowns and single unknown.
During a study session, a friend of yours absent-mindedly taps on his desk regularly at a rate of 11times every 41s.a) Determine the period of your friend's tapping.b) Determine the frequency of your friend's tapping.