Question

Don’t Miss the Boat. While on a visit to Minnesota (“Land of 10,000 Lakes”), you sign up to take an excursion around one of the larger lakes. Whe\({\bf{1500}} - {\bf{kg}}\)n you go to the dock where the boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude \({\bf{20}}\,{\bf{cm}}\). The boat takes \({\bf{3}}.{\bf{5}}\,{\bf{s}}\) 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 \({\bf{60}}\,{\bf{kg}}\)) begin to feel a bit woozy, due in part to the previous night’s dinner of lutefisk. As a result, you refuse to board the boat unless the level of the boat’s deck is within \({\bf{10}}\,{\bf{cm}}\) of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion.

Short answer

\(1.17\,{\rm{s}}\)

Step-by-step solution

Identification of given data

Mass of the boat \(m = 1500\,{\rm{kg}}\)

The amplitude \(A = \,\,{\rm{20}}\,{\rm{cm}}\)

Time period of boat \(T = 3.5\,{\rm{s}}\)

Significance of simple harmonic motion  

A motion known as a "simple harmonic motion," or SHM, is one in which the restoring force is directly proportional to the body's displacement from its mean position.

Displacement of SHM is expressed as,

\(\begin{aligned}{l}x = A\cos (\omega t + \varphi )\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,i.e\,\,\varphi = 0\\x = A\cos (\omega t)\,\end{aligned}\) …(i)

Where, \(x\) is the displacement, \(A\) is the displacement, \(\omega \) is the angular speed and \(t\) is the time

 Determining that how much time do you have to board the boat comfortably during each cycle of up-and-down motion 

Using equation (i)

\(x = A\cos (\omega t)\,\)

Substitute \(x = \frac{A}{2}\) and \(\omega = \frac{{2\pi }}{T}\) in above equation

\(\begin{aligned}{c}\frac{A}{2} = A\cos (\frac{{2\pi }}{T}t)\,\\\cos (\frac{{2\pi }}{T}t)\, = \frac{1}{2}\\\frac{{2\pi }}{T}t = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right)\\\frac{{2\pi }}{T}t = \frac{\pi }{3}\\t = \frac{T}{6}\end{aligned}\)

So the time taken for full up and down motion is given by,

\(\begin{aligned}{c}2t = \frac{T}{3}\\ = \frac{{3.5\,{\rm{s}}}}{3}\\ = 1.17\,{\rm{s}}\end{aligned}\)