Don't Miss the Boat: Get Onboard in 10cm or Less

[iwonde]

Homework Statement

While on a visit to Minnesota, you sign up to take an excursion around one of the larger lakes. When you go to the dock where the 1500-kg boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude 20cm. The boat takes 3.5 s to make one complete up-and-down cycle. When the boat is at its highest point, its deck is at the same height as the stationary dock. As you watch the boat bob up and down, you (mass 60 kg) begin to feel a bit woozy, due in part to the previous night's dinner. As a result, you refuse to board the boat unless the level of the boat's deck is within 10 cm of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion?

Homework Equations

The Attempt at a Solution

I don't really understand the problem. So am I finding the time intervals in which the level of the deck is within 10cm of the dock level? How should I approach this problem?

 

[Mapes]

A graph of simple harmonic motion looks like what function?

 

[qspeechc]

The motion has amplitude 20cm, you will only board if the boat is within 10cm of the dock, i.e. within 10 cm of its highest point in its motion. Draw a graph of the motion the boat executes and the answer should be easier to get.

 

[jghlee]

The motion has amplitude 20cm, you will only board if the boat is within 10cm of the dock, i.e. within 10 cm of its highest point in its motion. Draw a graph of the motion the boat executes and the answer should be easier to get.

I'm bringing back a really old question but it just happens that I have the same HW question...

I understand that the graph of simple harmonic motion is sin/cos, and the highest point corresponds to the 20 cm and then half of that will be 10 cm. But how do i figure out the time from that alone?

 

[cdotter]

Sorry to bump this thread, but how do I solve this problem? The amplitude is given as 0.2 m. The boat can only be boarded when it's within 0.1 m of the dock, so x = 0. T=2*Pi*sqrt(m/k), which means k must equal 4834 N/m, implying that omega is 1.795.

x = A*cos(omega*t)
0.1 = 0.2*cos(1.795*t)

How do I solve for t?

edit:

arccos(0.1/0.2)/1.795, but it gives 0.583 seconds. The answer is 1.17 seconds, or arccos(-0.1/0.2)/1.795. Why is it -0.1 (or -0.2, whichever it is)?