A particle of mass ,m, is at rest at the end of a spring(force constant=k)hanging from a fixed support. At t=0 a constant downward forcd F is applied to the mass and acts for a time T. Show that after the force is removed, the displacement of the mass from its equilibrium position (x=Xe, where x is down) is:
x - Xe =F/k[cosW(t-T) - cosw(t)]
where w = omega
Solution Summary
This solution shows that after the force is removed, the displacement of mass from equilibrium position is x - Xe =F/k[cosW(t-T) - cosw(t)]. Steps are shown for the full derivations for where the concepts of Hooke's Law and period are used.
... frequency of the oscillation b.) the period of the oscillation c.) the ... At each end of the rod are identical, ideal springs which have spring constants of ...
... Problem 13.6: The period of the spring as a harmonic oscillator is 2π ... m= × 0.2 = 5.48 N / m 0.6 0.6 So the springs force constant is ...
... Learning Goal: To derive the formulas for the major characteristics of motion as functions of time for a horizontal spring oscillator and to practice using ...
Simple harmonic oscillation driven by an external force. 2. A spring with a 4-kg mass has natural length 1 m and is maintained stretched to a length of 1.3 m by ...
... Restoring force developed in the spring is given by: F = - kx ___ _____ a) Frequency of oscillations is given by: ν = (1/2П)√k/m = (1/2П)√2460/7.66 ...
... the velocity of center of mass of the system thus the maximum extension in the spring can be calculated by conservation of energy of oscillation (relative to ...
... equation of motion for a damped oscillator, potential energy of identical springs, and the final resting position of a critically damped massless spring. ...
