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

Expert verified

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

Step by step solution

01

Identification of given data

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

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

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

02

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

03

 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,

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Question: The purity of gold can be tested by weighing it in air and in water. How? Do you think you could get away with making a fake gold brick by gold-plating some cheaper material?

You have probably noticed that the lower the tire pressure, the larger the contact area between the tire and the road. Why?

Given two vectors A=4.00i^+7.00j^ and B=5.00i^7.00j^, (a) find the magnitude of each vector; (b) use unit vectors to write an expression for the vector difference AB; and (c) find the magnitude and direction of the vector difference AB. (d) In a vector diagram showA,B and AB, and show that your diagram agrees qualitatively with your answer to part (c).

The most powerful engine available for the classic 1963 Chevrolet Corvette Sting Ray developed 360 horsepower and had a displacement of 327 cubic inches. Express this displacement in liters (L) by using only the conversions 1 L = 1000 cm3 and 1 in. = 2.54 cm.

A medical technician is trying to determine what percentage of a patient’s artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 1.20×104Pa, while in the region of blockage it is role="math" localid="1668168100834" 1.15×104Pa. Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 30.0 cm/s, and the specific gravity of this patient’s blood is 1.06. What percentage of the cross-sectional area of the patient’s artery is blocked by the plaque?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free