Respuesta :
Answer:
a) A = 0.98 m
b) Ф = 90°
c) x = -0.98sin(12.25t)
Explanation:
We know the value of the spring constant which is 300 N/m, the innitial apmplitude, that we will call it xo is 0, at t = 0, and the speed is 12 m/s
The expression for the amplitude under these conditions is:
A = √xo² + vx²/w² (1)
To calculate the angular speed w, we use the following expression:
w = √k/m (2)
Calculating w:
w = √300/2 = 12.25 rad/s
Now, we replace this value into equation 1, along with the other known values and solve for A:
A = √0 + (12)²/(12.25)²
A = 0.98 m
b) In this part, is actually easy, the displacement of x in function of the time is given by:
x = A cos(wt - Ф) (4)
But at t = 0 we have then:
x = xo = A cosФ (5)
Solving for the angle Ф we have:
xo/A = cosФ
Ф = arccos(x0/A) (6)
Replacing the data in (6):
Ф = arccos(0/0.98)
Ф = 90°
c) Equation (4) is the expression for the simple harmonic motion
x = A cos(wt - Ф)
And if we replace here the value of w and the previous angle, we can write an equation for x in function of t:
x = 0.98 cos (12.25t - 90)
x = 0.98 cos(12.25t - 90)
And we have an trigonometric expression for cos that is:
cos(α - π/2) = -sinα
in this case, α will be the value of w = 12.25 rad/s. and 90° is the same as rewritting as π/2, therefore:
cos(α - π/2) = cos(12.25t - 90)
x = -0.98sin(12.25t)
The amplitude is calculated as[tex]A = \sqrt{\frac{144}{150} }[/tex], The phase angle is [tex]\theta = \frac{\pi }{2}[/tex] and The equation for the position as a function of time is calculated as
[tex]x = -0.98sin(12.25t)[/tex]
Data;
- Mass = 2kg
- K = 300 N/m
- r = 12.0 m/s
Amplitude
The amplitude of the system can be calculated as
[tex]A = \sqrt{x_o^2 + \frac{v^2}{\omega^2} }\\x_o = 0\\\omega = \sqrt{\frac{k}{m} } = \frac{300}{2} \\ \omega^2 = 150[/tex]
Let's substitute this into the amplitude equation.
[tex]A = \sqrt{0+ \frac{12^2}{150} }\\A= \sqrt{\frac{144}{150} }[/tex]
The amplitude is calculated as
[tex]A = \sqrt{\frac{144}{150} }[/tex]
The Phase Angle
The phase angle of the system is calculated as
[tex]x = Acos(\omega t + \theta)\\t = 0\\x_o = Acos\theta\\0 = A cos\theta\\cos \theta = 0 \\\theta = \frac{\pi }{2}[/tex]
The phase angle is [tex]\theta = \frac{\pi }{2}[/tex]
Equation for the position as function of time
Using the standard equation for the function of time,
[tex]x = Acos(\omega t + \theta\\cos(\frac{\pi }{2} + \omega t) = -sin \omega t\\x = Acos(\omega t + \frac{\pi }{2})\\x = -0.98sin(12.25t)[/tex]
The equation for the position as a function of time is calculated as
[tex]x = -0.98sin(12.25t)[/tex]
Learn more on simple harmonic motion here;
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