Oscillation 2.0

Practice Problem:  A mass of 2 kg oscillating on a spring with constant 4 N/m passes through its equilibrium point with a velocity of 8 m/s. What is the energy of the system at this point? From your answer derive the maximum displacement, x m of the mass.

Solution:  K =.5 mv^2 =.5(2)(8)^2 = 64 joules  U =.5 kx^2  Since energy is conserved:  K =.5 mv^2 = U =.5 kx^2 = 64  X m = (x *2 / k) ^1/2 = 32^1/2 = 5.65 almost

Practiceeee  One end of alight spring with a spring constant 10 N/m is attached to a vertical support, while a mass is attached to the other end. The mass is pulled down and released, and exhibits simple harmonic motion with a period of.2 pie. The mass is .1 kg .25 kg .4 kg .04 kg .025 kg

Answer:  T= 2 pie (m/k)^1/2  M = kT^2 / 4 pie^2  = (10)(.2 pie)^2 / 4 pie ^2  M=.1 kg

Practice Problem  A mass m = 2.0 kg is attached to a spring having a force constant k = 290 N/m as in the figure. The mass is displaced from its equilibrium position and released. Its frequency of oscillation (in Hz) is approximately  a. 12  b  c  d. 1.9  e. 0.08

Solution: D, 1.9  W = (k/m)^.5  W = 2 pie f  W= (290 /2 )^.5 =  / 2 pie = f = 1.9 Hz

More Problems  Two circus clowns (each having a mass of 50 kg) swing on two flying trapezes (negligible mass, length 25 m) shown in the figure. At the peak of the swing, one grabs the other, and the two swing back to one platform. The time for the forward and return motion is  a. 10 s  b. 50 s  c. 15 s  d. 20 s  e. 25 s

Answer: A, 10 seconds  Is making 1 full cycle, so it is asking for T  T= 2 pie (L/g)^.5 =2 pie ( 25 / 9.81)^.5  Did it matter how much mass was on the rope??

Last Problem:  An enormous pendulum-driven clock, located on the earth, is set into motion by releasing its 10 meter long simple pendulum from a maximum angle of less than 10 degrees relative to the vertical. At what appreciated time t will the pendulum have fallen to a perfectly vertical orientation?  Pie / 10 seconds  Pie / 5 seconds  Pie / 4 seconds  Pie /2 seconds  Pie seconds

Solution: d, pie/2 seconds  T= 2 pie (L/g)^.5  T= 1/4T = ¼ * 2 pie (L/g)^.5  This is because we just want to know how long it takes to go back to the middle, which is ¼ an oscillation  T= 2 pie ( 10 m / 9.81 ( which we around to 10) ) ^.5  = 2 pie  Because 10 / 10 cancel out