Problem 1: A spring is hanging vertically from a support without any object atta
ID: 1879241 • Letter: P
Question
Problem 1: A spring is hanging vertically from a support without any object attached to it and its length is 500 mm. An object of mass 250 g is attached to the end of the spring. The length of the spring at equilibrium is 850 mm. The motion is described relatively to a vertical upward y-axis. The position of the mass relative to the relaxed position of the spring is noted "y (a) Prove that the position of equilibrium is determined by yo- What is the spring constant k? The spring is pulled down 120 mm and then released from rest. (b) Draw the free body diagram of the mass showing all forces applied on it. Air resistance is neglected. (c) Write the differential equation of the mass' displacement "u" relative to the position of equilibrium of mass-string system. Show that the motion is a simple harmonic motion. (Hint: we can write the position as y yo u) (d) What is the displacement amplitude? (e) What are the natural frequency of oscillation and period of motion? (f) What is the phase constant? (g) Write the equation of the motion u(t) (h) Show that the sum of kinetic energy, gravitational potential energy and elastic potential energy is constant. Calculate the total mechanical energy of the system. (Potential energies should be expressed by the position "y" relative to the relaxed position of the spring)Explanation / Answer
1) We Place the origin of our cordinate system at the top of the spring, with the positive direction upword, as is customary.
Then y0= -0.05m and y=-0.85m
The Spring Force is
fs= -k y0
Also, We know the force exerted on the spring was provided by the weight of the 0.25kg mass.
fs= -mg= -(0.25kg)(9.81m/s2)= - 2.4525N
Again the negative sign indicates the direction.
Now we can solve the force equation for the spring constant.
k= -fs/y0 = - 2.4525/-0.05 = 49.05 N/m.