Danielle Masso

Professor's note : All is good, to save your time and serve the purpose of this exercise, let's avoid the numbers. Give me the steps to solve a given problem in words - explain as much as you want !
--- keep up the good work !! ---

Week 2 (September 17, 2012)

Problem #1

m = 680g = 0.680kg
k = 65 N/m
t = 0
Xm/A = 12 cm = 0.12m
X = 0m

Part 1:

(angular frequency) w = sqrt(k/m) --> sqrt(65N/m / 0.680kg) = 9.7769

(period) T = (2pi) [sqrt(m/k)] --> (2pi)*[ sqrt(0.680kg / 65 N/m) ] = 0.64265 s

(frequency) f = 1/T --> 1 / 0.64265 s = 1.556 Hz


Part 2:

A = ?

X(t) = Acos(wt) --> 0.12m(t=0s) = Acos(9.7769 * 0s) --> 0.12m(t=0s) = Acos(0)

[cos(0)] = 1 --> 0.12m(t=0s) = A(1) --> 0.12m / 1 = 0.12m --> A = 0.12m



Part 3:

Vmax = A*w --> Vmax = 0.12m*9.7769 = 1.173228 m/s

Maximum speed occurs when the block is at the origin, x=0


Part 4:

a(max) = A*w^2 --> a(max) = 0.12m*(9.7769^2) = 11.47053 m/s^2


Part 5:

x(t) = Acos(wt)

0(t) = 0.12cos(9.7769t) --> 0(t) / 0.12 = [0.12cos(9.7769t)] / 0.12 --> 0 = cos(9.7769t) --> 9.7769t = cos^-1(0) --> 9.7769t = 1.57079 --> t = 0.160664 s


0.06(t) = 0.12cos(9.7769t) --> 0.06(t) / 0.12 = [0.12cos(9.7769t)] / 0.12 --> 0.5 = cos(9.7769t) --> 9.7769t = cos^-1(0.5) --> 9.7769t = 1.04719 --> t = 0.107109 s

(x=0m) - (x=0.06m) = ?
0.160664 - 0.107109 = 0.053555 s

Problem #2

(linear density) u = 7.2 g/m=0.0072 kg/m
(tension) F = 150 N
L = 90 cm = 0.9 m

Part 1:

v = sqrt(F / u) --> v = sqrt(150N / 0.0072 kg/m) = 144.3375 m/s

Part 2:

0.9m = 3lambda / 2 --> 2*0.9m = 3lambda --> 1.8m / 3 = 3lambda / 3 --> 0.6m = lambda


Part 3:

f = 1 / T

v = lambda / T --> T = lambda / v --> T = 0.6m / 144.3375 m/s = 0.0041569 s

--> f = 1 / 0.00415s = 240.5625 Hz

Week 3

Week 4