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Chapter 14: Problem 81

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. When you go to the dock where the \(1500 \mathrm{~kg}\) boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude \(20 \mathrm{~cm}\) The boat takes \(3.5 \mathrm{~s}\) 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 \(60 \mathrm{~kg}\) ) 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 \(10 \mathrm{~cm}\) 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
Therefore, you have about 0.914 seconds to board the boat comfortably during each cycle of up-and-down motion.

Step by step solution

01

Expression for Displacement in Simple Harmonic Motion

In simple harmonic motion, the displacement from the mean position can be given using the equation: \( d = A \cdot \cos(2\pi f t) \) Where, \( A \) is the amplitude of the motion, \( f \) is the frequency of the motion, \( t \) is the time, and \( d \) is the displacement from the mean position. The frequency \( f \) can be calculated as the reciprocal of the time period, which is 3.5 s in this case. That is, \( f = 1 / 3.5 = 0.286 \, \text{Hz} \).
02

Calculate Time When Boat's Deck is 10 cm of Dock Level

We are told that the boat deck need to be within 10 cm of the dock level to board the boat comfortably. Since the dock level corresponds to the highest point of the bobbing boat, it is safe to board when the boat's displacement is within 10 cm from its maximum amplitude. This means, looking at the equation : \( 10 = 20 \cdot \cos(2\pi f t) \) we get to \(0.5 = \cos(2\pi f t) \). Solving for \( t \), we get \( t = \cos^{-1}(0.5) / (2 \pi f) \) which approximately equals \( 0.457 \, \text{s} \)
03

Calculate Total Time to Board the Boat Comfortably

The time \( t = 0.457 \, \text{s} \) calculated in Step 2 is the time required for the boat to move from its highest point to a position where it is 10 cm within the dock level. This happens twice in one complete cycle - once when the boat is moving down from its highest point and other when it moving up towards its highest point. So total time to board comfortably is \( 2 \times 0.457 = 0.914 \, \text{s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Displacement
In simple harmonic motion, displacement is the measure of how far an object moves from its mean or equilibrium position. For the bobbing boat in our scenario, this movement is from the boat's highest point, which is on level with the dock, to its lowest during one cycle. The displacement in such cases is typically described using the equation:
\[ d = A \cdot \cos(2\pi f t) \]
Here, \(d\) represents the displacement at time \(t\), \(A\) is the amplitude, and \(f\) is the frequency. Notice that displacement is not the distance traveled, but simply how much away the object is from its central resting place. In the context of our problem, the displacement needed for the boat to be within safe boarding level is critical, emphasizing its role in understanding the motion.
Amplitude
Amplitude is a key parameter in simple harmonic motion, representing the maximum displacement from the equilibrium position. For our bobbing boat, this is the peak distance from the dock level to the highest point it reaches. An amplitude of 20 cm means that this is the furthest the boat moves from the mean level during its cycle.
  • The amplitude is always positive and reflects the strength of the wave motion.
  • It's the maximum height or depth in oscillating motion.
  • In practical terms, this tells us how much the boat will move up and down as it bobs on the water.
Understanding amplitude is crucial because it helps predict how far out of level a boat could be compared to the dock, dictating safe boarding times.
Frequency
Frequency is a fundamental concept when it comes to harmonic motion, dictating how often the oscillatory cycle occurs. It is calculated as the reciprocal of the period:
\[ f = \frac{1}{T} \]
where \(T\) is the period of one complete cycle. For our boat with a 3.5 s period, the frequency \(f\) is 0.286 Hz, meaning the boat completes nearly 0.286 cycles per second.
  • Frequency gives insight into how "quick" the motion is.
  • A higher frequency indicates a faster oscillation.
  • For this specific problem, knowing the frequency helps calculate how much time the boat is at a safe boarding level.
Frequency allows us to complehend the rhythm of the motion, which is pivotal in determining timings in dynamic situations.
Harmonic Motion
Simple harmonic motion (SHM) is characterized by oscillatory movement that follows a predictable, sinusoidal pattern. In the context of the problem, the boat's movement can be best described as such:
  • There is a restoring force directed toward the equilibrium position proportional to the displacement from it.
  • The movement is periodic, completing a cycle in a consistent time frame known as the period.
  • Amplitude, frequency, and displacement govern the characteristics and behavior of SHM.
Understanding SHM allows students to discern why the boat moves predictably, enabling informed decisions like boarding when the boat aligns closely with the dock.
Period of Oscillation
The period of oscillation is the time taken for the boat to complete one full cycle of motion. For the boat in our scenario, it is 3.5 seconds.
  • The period is directly related to frequency, as frequency is the inverse of the period.
  • A longer period means slower oscillation and vice versa.
  • In practical terms, it helps predict future displacement states.
The period is essential for solving when the motion achieves certain states, like alignment with the dock, highlighting its importance in applied physics and engineering scenarios.

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Most popular questions from this chapter

The Silently Ringing Bell. A large, \(34.0 \mathrm{~kg}\) bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of mass is \(0.60 \mathrm{~m}\) below the pivot. The bell's moment of inertia about an axis at the pivot is \(18.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The clapper is a small, \(1.8 \mathrm{~kg}\) mass attached to one end of a slender rod of length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length \(L\) of the clapper rod for the bell to ring silently - that is, for the period of oscillation for the bell to equal that of the clapper?

Object \(A\) has mass \(m_{A}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{A} .\) Object \(B\) has mass \(m_{B}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{B}\). The amplitude \(A_{A}\) for object \(A\) is twice the amplitude \(A_{B}\) for the motion of object \(B\). Also, \(m_{B}=4 m_{A}\) and \(k_{A}=9 k_{B}\). (a) What is the ratio of the maximum speeds of the two objects, \(v_{\max , A} / v_{\max , B} ?\) (b) What is the ratio of their maximum accelerations, \(a_{\max , A} / a_{\max , B} ?\)

A Spring with Mass. The preceding problems in this chapter have assumed that the springs had negligible mass. But of course no spring is completely massless. To find the effect of the spring's mass, consider a spring with mass \(M,\) equilibrium length \(L_{0},\) and spring constant \(k\). When stretched or compressed to a length \(L,\) the potential energy is \(\frac{1}{2} k x^{2},\) where \(x=L-L_{0}\). (a) Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\). Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of \(M\) and \(v .\) (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each piece in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L .\) The result is \(n o t \frac{1}{2} M v^{2},\) since not all of the spring moves with the same speed.) (b) Take the time derivative of the conservation of energy equation, Eq. (14.21), for a mass \(m\) moving on the end of a massless spring. By comparing your results to Eq. (14.8), which defines \(\omega\), show that the angular frequency of oscillation is \(\omega=\sqrt{k / m}\). (c) Apply the procedure of part (b) to obtain the angular frequency of oscillation \(\omega\) of the spring considered in part (a). If the effective mass \(M^{\prime}\) of the spring is defined by \(\omega=\sqrt{k / M^{\prime}},\) what is \(M^{\prime}\) in terms of \(M ?\)

A \(175 \mathrm{~g}\) glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant \(155 \mathrm{~N} / \mathrm{m}\). At the instant you make measurements on the glider, it is moving at \(0.815 \mathrm{~m} / \mathrm{s}\) and is \(3.00 \mathrm{~cm}\) from its equilibrium point. Use energy conservation to find (a) the amplitude of the motion and (b) the maximum speed of the glider. (c) What is the angular frequency of the oscillations?

A \(0.500 \mathrm{~kg}\) mass on a spring has velocity as a function of time given by \(v_{x}(t)=-(3.60 \mathrm{~cm} / \mathrm{s}) \sin [(4.71 \mathrm{rad} / \mathrm{s}) t-(\pi / 2)] .\) What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?
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