Chapter 4, Problem 145CR

To determine

To Compute : An equation representing the motion of the buoy.

Answer to Problem 145CR

The equation representing the distance travelled by the buoy is, $y=3\cos (\frac{2\Pi }{15}t)$

Explanation of Solution

Given information : The buoy is oscillating in a simple harmonic motion. The total distance travelled by the buoy vertically is 6 ft. And the high point corresponds to time, t = 0.

As the high point of the buoy corresponds to time t = 0, the function representing the distance travelled by buoy will be a cosine function.

That is,

  $y=A\cos (\omega t)$ ........(1)

Where, A represents the amplitude, $\omega$ represents the angular velocity of the buoy and T is the time period and t is the time.

Since amplitude A of a simple harmonic motion represents the vertical displacement towards both the sides. Therefore the vertical distance from high to low point covers twice the amplitude.

Therefore, A = 3 feet.

As time period T = 15 sec, the angular velocity is,

  $\omega =\frac{2\Pi }{T}=\frac{2\Pi }{15}$ .

By substituting $\omega$ and A in equation number (1), implies,

  $y=3\cos (\frac{2\Pi }{15}t)$ .

Thus,

The equation representing the distance travelled by the buoy is,

  $y=3\cos (\frac{2\Pi }{15}t)$