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Chapter 16: Problem 15

15\. Harbor At a certain harbor, the tides cause the ocean surface to rise and fall a distance \(d\) (from highest level to lowest level) in simple harmonic motion, with a period of \(12.5 \mathrm{~h} .\) How long does it take for the water to fall a distance \(d / 4\) from its highest level?

Short Answer

Expert verified
The time required is approximately 2.08 hours.

Step by step solution

01

Identify the given values

The period of the harmonic motion is given as 12.5 hours. This is the time it takes to complete one full cycle of rising and falling. The maximum amplitude is the distance from the highest to the lowest level, denoted as distance d.
02

Understand the properties of simple harmonic motion

In simple harmonic motion, the displacement over time can be described by sine or cosine functions. The general equation for the height can be expressed as: \[ h(t) = A \text{cos}\bigg(\frac{2\pi}{T}t + \theta_0\bigg) \]where: - \( A \) is the amplitude (\( d/2 \)), - \( T \) is the period (12.5 hours), - \( t \) is time, - \( \theta_0 \) is the phase constant.
03

Set up for the specific time to fall distance \( d / 4 \) from highest level

The highest level corresponds to maximum amplitude, so the height is initially \( d/2 \). We want the height to be \( d/4 \) less than this, so the target height is:\[ h(t) = \frac{d}{2} - \frac{d}{4} = \frac{d}{4} \]
04

Solve the cosine equation for time

Using the height function:\[ \frac{d}{4} = \frac{d}{2} \text{cos}\bigg(\frac{2\pi}{12.5}t \bigg) \]We simplify and solve:\[ \text{cos}\bigg(\frac{2\pi}{12.5}t \bigg)= \frac{1}{2} \]\[ \frac{2\pi}{12.5}t = \frac{\pi}{3} \]\[ t = \frac{12.5}{6} \approx 2.08 \text{ hours} \]

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Key Concepts

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

periodic motion
Periodic motion is a type of motion that repeats itself at regular intervals. Think of it like the ticking of a clock 鈥� it's consistent and predictable. In terms of tides, it means the rise and fall of the water at the harbor happen in a cyclical pattern. Each cycle of the tide鈥檚 high and low points takes the same amount of time, known as the period. For the harbor problem, this period is 12.5 hours. This regularity is what characterizes periodic motion, making it easier to predict and calculate various points in time, such as how long it takes for the water to fall a certain distance.
amplitude
Amplitude is the maximum extent of displacement from the equilibrium or center position. In the context of tides, it is half the distance between the highest and lowest tide levels. If we denote the entire rise and fall distance as 'd', then the amplitude, which is the maximum distance from the highest point to the middle of the oscillation, would be \(\frac{d}{2}\). The amplitude helps us understand how far the water levels can go above or below the average (middle) point. For our problem, knowing the amplitude is crucial as it feeds directly into the formula we use to solve for displacement over time.
cosine function
In simple harmonic motion (SHM), the cosine function is often used to describe how a quantity changes over time. The cosine function is particularly helpful because it starts at its maximum value when time is zero, which corresponds to the highest tide level in our harbor problem. The general equation used is: \[ h(t) = A \cos\bigg(\frac{2\pi}{T}t + \theta_0\bigg) \] Here, 'A' is the amplitude, 'T' is the period, 't' is time, and \(\theta_0\) is the phase constant. The cosine function shows how the height 'h' changes over time, following a smooth, wave-like pattern, making it ideal for modeling tides.
displacement over time
Displacement over time is about how the position of something (like water in our case) changes as time progresses. For simple harmonic motion, the displacement is described by our cosine function. When dealing with tides, displacement tells us how high or low the water is at a specific time. In our problem, we wanted to find the time it takes for the water level to fall from its highest point down a distance of \(\frac{d}{4}\). By setting up and solving the equation \[ \frac{d}{4} = \frac{d}{2} \cos\bigg(\frac{2\pi}{12.5}t \bigg) \] we can find the exact time of 2.08 hours. This tells us how displacement varies in direct relation to time within the cycle.

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Most popular questions from this chapter

3\. Oscillator An oscillator consists of a block of mass \(0.500 \mathrm{~kg}\) connected to a spring. When set into oscillation with amplitude \(35.0 \mathrm{~cm}\), the oscillator repeats its motion every \(0.500 \mathrm{~s}\). Find (a) the period, (b) the frequency, (c) the angular frequency, (d) the spring constant, (e) the maximum speed, and (f) the magnitude of the maximum force on the block from the spring.

31\. Balance Wheel The balance wheel of a watch oscillates with a rotational amplitude of \(\pi\) rad and a period of \(0.500 \mathrm{~s}\) Find (a) the maximum rotational speed of the wheel, (b) the rotational speed of the wheel when its displacement is \(\pi / 2 \mathrm{rad}\), and (c) the magnitude of the rotational acceleration of the wheel when its displacement is \(\pi / 4 \mathrm{rad}\).

19\. Oscillator An oscillator consists of a block attached to a spring \((k=400 \mathrm{~N} / \mathrm{m}) .\) At some time \(t\), the position (measured from the system's equilibrium location), velocity, and acceleration of the block are \(x=0.100 \mathrm{~m}, v=-13.6 \mathrm{~m} / \mathrm{s}\), and \(a=-123 \mathrm{~m} / \mathrm{s}^{2} .\) Calcu- late (a) the frequency of oscillation, (b) the mass of the block, and (c) the amplitude of the motion.

44\. In a Car A simple pendulum of length \(L\) and mass \(m\) is suspended in a car that is traveling with constant speed \(|\vec{v}|\) around a circle of radius \(R\). If the pendulum undergoes small oscillations in a radial direction about its equilibrium position, what will be its frequency of oscillation?

11\. Automobile Spring An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of \(3.00 \mathrm{~Hz}\). (a) What is the spring constant of each spring if the mass of the car is \(1450 \mathrm{~kg}\) and the mass is evenly distributed over the springs? (b) What will be the oscillation frequency if five passengers, averaging \(73.0 \mathrm{~kg}\) each, ride in the car? (Again, consider an even distribution of mass.)
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