Question
c. find the appeoximate value foe rwhen t-2 SECONDS (pes)? CAREFUL d. Ifthe value of the constans Eo were to triple (while all else remained fixed), then wwbat wonld bappen to the mamber ofoscillations we would expect to count per bow? CAREFUL (Would the number increase or decrease, by precisely how much) pts.) Page of 1o b. (includes all four parts below) (35 pts). Near the surface of Earth, a particle of mass m is dangled from a long string, length L; the particle oscillates along a small are according to the differential equation where refers to an angular displacement measured from the vertical and t refers to The particle's mass is given by m = 8 kg The length of the string, L, isconstant and was accidentally not recorded by the researchers-but can be deduced from all the other given information. Whenever the particle arrives at a location of ,, 2 radians from the vertical, the particle has no instantaneous speed. On both sides of the vertical, that is, 2 radians is repeatedly observed to be a turning point for the particle's periodic motion i. Draw a clear diagram of this particle at some arbitrary point during oscillation, making sure to label variables and constants described above (4 pts). ii. In what nits should the constant (20) be measured (3 pts)? ii. Approximating to three significant digits, what is the angular frequency of this oscillator on a string (4 pts)? iv. What is tbe particle's approximate SPEED at t = 3 seconds (4 pts V. Assume that an experimenter begins measuring time at the instant the particle reaches 2 radians from the vertical. Assume, further, that the only force doing work on this dangling particle is gravity How mucb Potential Energy does the particle have at t = 0- immediately at the start (0 pts)? ) (b) Assume that all the above values remain precisely the same, EXCEPT that the turning points at +/ 2 radians now occur at t/6 radians: In other words, the amplitude has been TRIPL.ED. Given this particular change, describe what will happen to the value of the period of time for one cycle (10 pts)
Explanation / Answer
Page 3 of 10
given x = e^-wt
then dx/dt = -we^-wt
d^2x/dt^2 = w^2e^-wt
Page 4 of 10
i. let x = Acos(wt)
then
dx/dt = -Awsin(wt)
d^x/dt^2 = -Aw^2cos(wt)
ii. given equation d^2x/dt^2 = -w^2*x
putting in values
-Aw^2cos(wt) = -w^2*Acos(wt)
LHS = RHS
hence x = Acos(wt) is a solution to the given equation
a) xa = -ie^-iwt
i. dx/dt = i^2*e^(-iwt) = -e^(-iwt)
ii. d^2x/dt^2 = ie^(-iwt)
b) z = zo^2sin(wt + i^3) = zo^2sin(wt - i)
i. dz/dt = zo^2*w*cos(wt - i)
ii. d^2z/dt^2 = -zo^2*w^2*sin(wt - i)
c) given equation
d^2x/dt^2 = -w^2x
-zo^2*w^2*sin(wt - i) = -w^2*zo^2sin(wt - i)
LHS = RHS
hence x = zo^2sin(wt - i) is the solution of the given eqwuation
