Consider a simple harmonic system consisting of a 2.5 kg mass and a nearly mass-

Question

Consider a simple harmonic system consisting of a 2.5 kg mass and a nearly mass-less spring hung vertically. When the mass is attached to the bottom of the spring it elongates the spring by 1.75 cm. (a) Find the spring constant of the spring. (b) If you pull the mass downward from equilibrium a distance of 1.5 cm and release it you will set it into oscillation. In this case, find the angular frequency of the system. (c) For the oscillating system of part (b) write the equation of motion for this oscillator in the form y(t) = ymaxCos(?t). (d) What is the velocity of the mass at t = 2 seconds? Answer: (a) 1400 N/m (b) ? = 23.7 rad/sec (c) y=0.015cos23.7t (d) 0.097m/s

Explanation / Answer

a)spring constant=2.5*9.8/(1.75*0.01)=1400 N/m b)angular frequency=sqrt(spring constant/mass)=sqrt(1400/2.5)=23.6643 /sec c)y(t)=0.015*cos(23.6643*t) m d)speed=y'(t)=0.015*sin(23.6643*2)*23.6643=0.2617 m/s

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