Respuesta :
Answer:
A. 1.77m
B. -33.3m/s
C. -628.27m/s²
D. 2π² or 26°
E. 3Hz
F. 0.33s
Explanation:
Given X = 2.5cos(6πt + π/4)
(a) at t = 7
X = 2.5cos[6π(7) +π/4)
= 2.5cos[42π+π/4] = 2.5cos(169π/4)
= 2.5cos[21×360+ 45]
= 2.5cos(π/4)
= 1.7677m
~=1.77m
Note: π =180°
Hence 169π/4 = 7605°= [21×360 +45] =45°
(B) velocity = dX/dt
dX/dt = -2.5sin(6πt +π/4) ×6π........
= -15πsin[6π(7)+π/4]
= -15πsin(169π/4)
= -15πsin(π/4)
= -33.325
~= -33.3m/s
(C) acceleration; a = d²x/dt² = X"
x" = d(dx/dt)/dt
x" = -15sin(6πt+π/4) × 6π
= -90π²cos(6π+ π/4)
= -90π²cos(π/4)
= - 628.26m/s²
(D) phase angle = wπT
= (2πf)π ×1/f
= 2π² = 180π = 566° = 360+206
= 206 = 180° +26°
= 26°
Note π=180°
(E) using the acceleration, a we use the formula:
a = - w²x
w = 2πf
a = - (2πf)²x
a = -4π²f²x
f = √(a/4π²x)= 1/(2π)√(a/x)
= 0.1591√(628.26/1.77)
= 2.998
~= 3Hz
At t= 7.0, x= 1.77m
(F) T = 1/f = 1/2.998
T = 0.3335s
Answer:
Explanation:
Given:
Displacement, x = (2.5 m) cos[(6π rad/s)t + π/4 rad]
A.
At t = 7s,
x = 2.5 × cos(42π + π/4)
= 2.5 × cos(169/4 × π)
= 5/4 × sqrt2
= 1.77 m
B.
dx/dt = v = -(2.5 × 6π) × sin[(6π rad/s)t + π/4 rad]
= -15π × sin[(6π rad/s)t + π/4 rad]
At t = 7s,
= -15π × sin[(42π rad/s)t + π/4 rad]
= -15π × sin(169/4 × π)
= -15/2 × π × sqrt2
= -33.32 m/s
C.
dv/dt = a = -(2.5 × (6π)^2) × cos[(6π rad/s)t + π/4 rad]
= -90 × (π)^2) × cos[(6π rad/s)t + π/4 rad]
At t = 7s,
= -90 × (π)^2) × cos[(42π rad/s) + π/4 rad]
= -45 × (π)^2) × sqrt2
= -628.1 m/s^2
D.
Comparing ,
x = Acos(wt + phil)
With,
x = (2.5 m) cos[(6π rad/s)t + π/4 rad]
Phase angle, phil = π/4 rad
Since 2π rad = 360°
π/4 rad = 360/8
= 45°
E.
angular velocity, w = 2π/t
= 2π × f
Comparing the above equations,
w = 6π rad/s
Frequency, f = 6π/2π
= 3 Hz
F.
Period, t = 1/f
= 1/3
= 0.33 s.
