Respuesta :
Answer:
a) [tex]\omega = 10.407\,\frac{rad}{s}[/tex], b) [tex]m = 4.617\,kg[/tex], c) [tex]A = 1.355\,m[/tex]
Explanation:
a) The system have a simple armonic motion, whose position function is:
[tex]x(t) = A\cdot \cos (\omega\cdot t + \phi)[/tex]
The velocity function is determined by deriving the position function in terms of time:
[tex]v(t) = -\omega \cdot A \cdot \sin(\omega\cdot t + \phi)[/tex]
The acceleration function is found by deriving again:
[tex]a(t) = -\omega^{2} \cdot A \cdot \cos (\omega\cdot t + \phi)[/tex]
Let assume that [tex]t = 0\,s[/tex]. The following nonlinear system is built:
[tex]A\cdot \cos \phi = 0.660\,m[/tex]
[tex]-\omega \cdot A \cdot \sin \phi = -12.3\,\frac{m}{s}[/tex]
[tex]-\omega^{2}\cdot A \cdot \sin \phi = -128\,\frac{m}{s^{2}}[/tex]
System can be reduced by divinding the second and third expressions by the first expression:
[tex]\omega \cdot \tan \phi = 18.636\,\frac{1}{s}[/tex]
[tex]\omega^{2}\cdot \tan \phi = 193.94\,\frac{1}{s^{2}}[/tex]
Now, the last expression is divided by the first one:
[tex]\omega = 10.407\,\frac{rad}{s}[/tex]
b) The mass of the block is:
[tex]m = \frac{k}{\omega^{2}}[/tex]
[tex]m = \frac{500\,\frac{N}{m} }{(10.407\,\frac{rad}{s})^{2} }[/tex]
[tex]m = 4.617\,kg[/tex]
c) The phase angle is:
[tex]\phi = \tan^{-1} \left(\frac{18.636\,\frac{1}{s} }{\omega} \right)[/tex]
[tex]\phi \approx 0.338\pi[/tex]
The amplitude is:
[tex]A = \frac{0.660\,m}{\cos 0.338\pi}[/tex]
[tex]A = 1.355\,m[/tex]
Answer:
Explanation:
Given:
Spring constant, k = 500 N/m
Displacement, x = 0.660 m
Velocity, v = -12.3 m/s
Acceleration, a = -128 m/s2
For a body experiencing simple harmonic motion,
x = A cos (ωt + φ)
0.66 = A cos (ωt + φ) ....1
dx/dt = A cos (ω t + φ ) dt
dx/dt = v = -Aω × sin (ω t + φ)
-12.3 = -Aω × sin (ω t + φ) ......2
dv/dt = -Aω × sin (ω t + φ) dt
dv/dt = a = -Aω^2 × cos (ω t + φ)
-128 = -Aω^2 × cos (ω t + φ) .......3
Equating equation 1 and 3,
-128 = -ω^2 × 0.66
ω^2 = 128/0.66
= 193.94
ω = 13.93 rad/s
ω = 2pi × f
frequency, f = 13.93/2pi
= 2.22 Hz
B.
Using Hooke's law,
Force, F = -kx
Force = mass, m × acceleration, a
Mass = (500 × 0.66)/128
= 2.58 kg
C.
Amplitude, A
ω = 13.93 rad/s
Frome equation 2 and 3,
-12.3 = -Aω × sin (ω t + φ)
-12.3 = -A × 13.93 × sin (13.93 × 1/2.22 + φ)
0.883 = A × sin (6.275 + φ) .....4
-128 = -Aω^2 × cos (ω t + φ)
-128 = A (13.93)^2 cos (13.93 × 1/2.22 + φ)
0.66 = A cos (6.275 + φ) .....5
From equation 4 and 5,
0.883 = A × sin (6.275 + φ)
0.66 = A cos (6.275 + φ)
Squaring both and equating them,
0.78/A^2 = sin^2 (6.275 + φ)
0.436/A^2 = cos^2 (6.275 + φ)
Adding both,
0.78/A^2 + 0.436/A^2 = sin^2 (6.275 + φ) + cos^2 (6.275 + φ)
From sin^2 theta + cos^2 theta = 1
0.78/A^2 + 0.436/A^2 = 1
0.78 + 0.436 = A^2
A = sqrt(1.2156)
= 1.1025 m
