Mathematics for Construction

Running head: MATHEMATICS FOR CONSTRUCTION
MATHEMATICS FOR CONSTRUCTION
Name of the Student
Name of the University
Author Note

MATHEMATICS FOR CONSTRUCTION1
Scenario 1:
Given equations of vibration,
x1=3.25 sin(100 πt + 2 π
9 ¿)¿
x2=4.44 sin(100 πt− 2 π
5 ¿) ¿
i. amplitude of x1 = 3.25, amplitude of x2 = 4.44
phase of x1 = 2 π
9 ,phase of x2 = −2 π
5
frequency of x1 = 100 π /2 π = 50 Hz.
Hence, time period of x1 = 1/50 = 0.02 secs.
frequency of x2 = 100 π /2 π = 50 Hz.
Hence, time period of x2 = 1/50 = 0.02 secs.
ii. Maximum displacement of x1 = 3.25
Hence, 3.25 sin(100 πt+ 2 π
9 ¿)=3.25 ¿
 sin(100 πt + 2 π
9 ¿)=1 ¿
 100πt + 2 π
9 = arcsin(1)
 100πt + 2 π
9 = π/2
 100 πt = 5π/18
 t = 5/1800 = 0.0027 secs.

MATHEMATICS FOR CONSTRUCTION2
Hence, maximum amplitude of vibration is produced from machine 1 at time t = 0.0027 secs.
Similarly,
4.44 sin(100 πt− 2 π
5 ¿)=4.44 ¿
 100 πt = π
2 + 2 π
5 = 9 π
10
 t = 9/1000 = 0.009 secs.
Hence, the vibration of machine 2 reaches maximum amplitude at time t = 0.009 secs.
iii. The time at which the vibration reaches to a displacement of -2 mm is found by the
following equations.
x1=3.25 sin(100 πt + 2 π
9 ¿)¿ = -2
x2=4.44 sin(100 πt− 2 π
5 ¿) ¿ = -2
3.25 sin(100 πt + 2 π
9 ¿)¿ = -2
 sin(100 πt + 2 π
9 ¿)¿ = -2/3.25
 100 πt + 2 π
9 = arcsin(-2/3.25)
 100 πt + 2 π
9 = π + 0.66287
 100 πt = 3.1063
 t=¿0.00988 secs.
Hence, vibration of machine 1 reaches -2 mm of displacement at time t = 0.00988 secs.
4.44 sin(100 πt− 2 π
5 ¿)¿ = -2

MATHEMATICS FOR CONSTRUCTION3
 sin(100 πt − 2 π
5 ¿)¿ = -2/4.44
 100 πt− 2 π
5 = arcsin(-2/4.44)
 100 πt = 4.8655
 t=¿0.01548 secs.
Hence, vibration of machine 2 reaches -2 mm at time t = 0.01548 secs.
iv. x1=3.25 sin(100 πt+ 2 π
9 ¿)¿
= 3.25(sin(100 πt)cos( 2 π
9 ¿+ cos ( 100 πt )∗sin ( 2 π
9 )¿
= 2.49sin(100πt) + 2.089cos ( 100 πt )
Hence, for x1 A = 2.49, B = 2.089
Similarly,
x2=4.44 sin(100 πt− 2 π
5 ¿) ¿
= 4.44(sin(100 πt ¿ cos (2 π
5 ) – cos(100 πt ¿∗sin ( 2 π
5 )¿
= 1.372sin(100 πt ¿ – 4.223cos(100 πt ¿
Hence, for x2 A = 1.372 and B = -4.223
v. Now, from the above result
x1 + x2 = 2.49sin(100πt) + 2.089cos ( 100 πt ) + 1.372sin(100 πt ¿ – 4.223cos(100 πt ¿
= 3.862sin(100πt) – 2.134cos(100 πt ¿
Rcos(theta) = 3.862 and Rsin(theta) = -2.134
tan(theta) = -2.134/3.862 = -0.55256
 theta = arctan(-0.55256) = -0.50481 rad
Hence, R = 3.862/cos(-0.50481) = 4.41237
Hence, x1 + x2 = 4.41237sin(100 πt−0.50481¿