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)
