answers and explanations please

QUESTIONS 1. Calculate the phase angle differences between position (displacement) and velocity(AQpv), position and acceleration(AQpa), and velocity and acceleration(AQva) using the data from Data Table 1 and 2. App = Pp - Qv Appa = Do - Da Adva = Qv - Da 2. If the displacement x(t) of a mass-spring system is given by x(t) = A . cos(w-t - ( ) = A . cos(2xf.t - (), show that Vmax = Aw = A- [k/m]12 and amax = Aw- = A. [k/m]. 3. Show that the free fall acceleration g is given by g = yo (2n/T) where yo is the stretched length of the spring caused by a suspended mass m, T is the period, and @ = (2n/T) = (2xf)= = k/m. 4. (a) If we neglect the spring's mass, is the experimental value k (based on the equation k = my/yo = m. ) less or bigger than actual value of k? and (b) why equation g = yo (2n/T) is a better approximation for a large mass m.Part A: Measuring A. T. f. Q. Umm, and aman- Data Table I: Position Graph Measurement of Amplitude A. Period T. Frequency f, and Position Phase Angle O ym) y(m) Aim) N TIS fi Hz) (ofrad) 0.502 0.395 052 0.05 4.15 B 0.586 1.71 0.536 Data Table 2: Velocity and Acceleration Graphs Measurement of Phase Angles of Velocity O, and Acceleration ., Maximum Speed Um. and Acceleration as @. rad D.(rad) Un(m/'s) 0.5 5.36 0.35 3.758-3 0.52 5.4 Data Table 3: Calculation of Maximum Speed Um and Acceleration am % Difference" "% Difference"* 0.558 7.05 5.98 10.19 "A and T are from Data Table 1. * % Difference is between Data Table 2 and 3. Part B: Measuring Spring Constant k and Free Fall Acceleration g Data Table 4: Position Graphs with Varying Masses Bottom position of unstressed spring with respect to the floor: ha = 1.34 Mikg) him) ye (m) N T(s) (m/s) 1/M(L/kg) 0.040 1.285 0.055 0.1 5.35 11 0.525 7.87 25 0.050 1.125 0.125 0.45 3.45 0.5 19.72 20 0.070 1.18 0.16 0.50 6 0.68 13.65 14.29 0.090 1.145 0.195 0.15 3 95 0.76 13.31 11.11 0.100 1.13 0.21 0.75 5.55 0.B 12.94 10