Problem 23
Question
An object moves in simple harmonic motion described by the given equation, where \(t\) is measured in seconds and \(d\) in inches. In each exercise, find the following: a. the maximum displacement b. the frequency c. the time required for one cycle. $$ d=-6 \cos 2 \pi t $$
Step-by-Step Solution
Verified Answer
The maximum displacement is 6 inches, the frequency is 1 Hz, and the time required for one cycle is 1 second.
1Step 1: Find the Maximum Displacement
In the equation \(d=-6 \cos(2 \pi t)\), the coefficient of the cosine function is -6. This represents the maximum displacement of the object from its equilibrium position, also known as the amplitude. Since displacement can't be negative, we take the absolute value giving us \(A =| -6 |= 6\) inches.
2Step 2: Find the Frequency
The frequency can be obtained from the coefficient of \(t\) in the equation. It is given by \(2 \pi\), which represents the angular frequency \(\omega\). The frequency \(f\) can be found by dividing the angular frequency \(\omega\) by \(2 \pi\). This gives us \(f = \frac{2 \pi}{2 \pi} = 1\) Hz.
3Step 3: Find the Time required for One Cycle
The time required for one complete cycle, also known as the period \(T\), is the reciprocal of the frequency. So, \(T = \frac{1}{f} = \frac{1}{1} = 1\) second.
Key Concepts
Maximum DisplacementFrequency of OscillationPeriod of Motion
Maximum Displacement
When discussing simple harmonic motion (SHM), one critical parameter is the 'maximum displacement' which represents how far an object moves from its equilibrium, or rest position, at the peak of its oscillation. In our exercise, we are given the equation
In this scenario, we identify the amplitude by taking the absolute value to get \(A = 6\) inches. This distance is also known as the 'peak amplitude,' since it describes how far from the central position the object will travel during its oscillation. It is crucial to understand that the amplitude plays no role in the oscillation's frequency or period; it only reflects the extent of the motion.
d = -6 \cos(2 \pi t), where d represents the displacement in inches, and t is the time in seconds. The number in front of the cosine function, in this case, -6, is known as the amplitude of the wave. The amplitude is the absolute value of the maximum displacement, meaning it's always a positive quantity regardless of the equation's negative sign.In this scenario, we identify the amplitude by taking the absolute value to get \(A = 6\) inches. This distance is also known as the 'peak amplitude,' since it describes how far from the central position the object will travel during its oscillation. It is crucial to understand that the amplitude plays no role in the oscillation's frequency or period; it only reflects the extent of the motion.
Frequency of Oscillation
The 'frequency of oscillation' is another vital aspect of SHM and it determines how often the oscillations occur within a specific time period. Technically speaking, it is the number of complete cycles that occur per unit time and is measured in hertz (Hz), where 1 Hz is equal to one cycle per second. In the given exercise, our equation is
To find the regular frequency \(f\) from angular frequency, we divide \(\omega\) by \(2 \pi\), yielding \(f = \frac{2 \pi}{2 \pi} = 1\) Hz. This result indicates that our object completes one oscillation cycle each second. It's essential to grasp that higher frequency means more cycles per second, which in turn signifies faster oscillation.
d = -6 \cos(2 \pi t). Here, the angular frequency \(\omega\) comes from the coefficient of t, which is \(2 \pi\).To find the regular frequency \(f\) from angular frequency, we divide \(\omega\) by \(2 \pi\), yielding \(f = \frac{2 \pi}{2 \pi} = 1\) Hz. This result indicates that our object completes one oscillation cycle each second. It's essential to grasp that higher frequency means more cycles per second, which in turn signifies faster oscillation.
Period of Motion
Lastly, we consider the 'period of motion' which is intrinsically connected to frequency. The period, denoted by \(T\), is the amount of time it takes to complete one full cycle of oscillation. It can be thought of as the temporal duration of one oscillation. For simple harmonic motion, this concept becomes particularly important as SHM is defined by this repetitive, cyclical nature. To find the period from frequency, we use the relationship that the period is the reciprocal of the frequency.
In our exercise, the frequency is 1 Hz as previously determined, so calculating the period is straightforward,
In our exercise, the frequency is 1 Hz as previously determined, so calculating the period is straightforward,
T = \frac{1}{f} = \frac{1}{1} = 1 second. This concise time period tells us the object takes exactly one second to return to its starting point in its motion path. Understanding the period is essential for predicting when an object in SHM will be at a particular point in its oscillation cycle.Other exercises in this chapter
Problem 22
In Exercises \(21-28,\) convert each angle in radians to degrees. $$ \frac{\pi}{9} $$
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Use a calculator to find the value of each expression rounded to two decimal places. $$ \cos ^{-1} \frac{3}{8} $$
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Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$y=\frac{1}{2} \sin \left(x+\frac{\pi}{2}\right)$$
View solution Problem 23
In Exercises 17–24, graph two periods of the given cotangent function. $$ y=3 \cot \left(x+\frac{\pi}{2}\right) $$
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