Question

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

Answer 1

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)