Meagan is sitting in a rocking chair, the distance d(t-) between the wall and the rear of the chair varies sinusoidally with time t. At t = 1s , the chair is closest to the wall and d(1) = 18 cm. At t = 1.75s the chair is farthest from the wall and d(1.75 ) = 34m

Question

how far is the chair when no one is rocking it?

and

what is the distance between wall and the chair at t= 8s?

Answer 1

Let's use the equation d(t) = A * sin(B * (t - C)) + D. The amplitude A is equal to half the difference between the maximum and minimum distances (34 cm - 18 cm) / 2 = 16 cm / 2 = 8 cm. The center of oscillation D is the average of the maximum and minimum distances, (34 cm + 18 cm) / 2 = 52 cm / 2 = 26 cm.

Now we can find the phase shift C. Since Meagan is closest to the wall at t = 1s, we have sin(B * (1 - C)) = -1, because the sine function reaches its minimum at a 270 degrees or 3π/2 radians, we can write the equation as:

B * (1 - C) = 3π/2

We can also find the period of the oscillation using the information given about the chair at t = 1.75s. Since the chair is farthest from the wall at t = 1.75s, we have sin(B * (1.75 - C)) = 1. We know that the sine function reaches its maximum at 90 degrees or π/2 radians, so we can write the equation as:

B * (1.75 - C) = π/2

Since both equations equal the sine function's maximum and minimum, respectively, we can set them equal to each other:

3π/2 * (1 - C) = π/2 * (1.75 - C)
3(1 - C) = 1.75 - C
C = 2/7

Now we can find B:

B * (1 - 2/7) = 3π/2
B * (5/7) = 3π/2
B = (3π/2) * (7/5)
B = 3π/2 * 7/5

We're now ready to find the distance between the wall and the chair when no one is rocking it, which is at t = 0:

d(0) = 8 * sin(3π/2 * 7/5 * (0 - 2/7)) + 26
d(0) = 8 * sin(-6π/5) + 26
d(0) ≈ 8 * 0.951 + 26
d(0) ≈ 7.608 + 26
d(0) ≈ 33.608 cm (approximately)

So when no one is rocking the chair, the distance between the wall and the chair is about 33.608 cm.

Now let's find the distance between the wall and the chair at t = 8s:

d(8) = 8 * sin(3π/2 * 7/5 * (8 - 2/7)) + 26
d(8) ≈ 8 * sin(0.760) + 26
d(8) ≈ 8 * 0.686 + 26
d(8) ≈ 5.488 + 26
d(8) ≈ 31.488 cm (approximately)

So the distance between the wall and the chair at t = 8s is about 31.488 cm.