/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 89 A weight is attached to a spring... [FREE SOLUTION] | 91Ó°ÊÓ

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A weight is attached to a spring suspended vertically from a ceiling. When a driving force is applied to the system, the weight moves vertically from its equilibrium position, and this motion is modeled by \( y = \dfrac{1}{3} \sin 2t + \dfrac{1}{4} \cos 2t \) where \( y \) is the distance from equilibrium (in feet) and \( t \) is the time (in seconds). (a) Use the identity \( a \sin B \theta + b \cos B \theta = \sqrt{a^2 + b^2} \sin (B\theta + C) \) where \( C = \arctan (b/a) \), \( a > 0 \), to write the model in the form \( y = \sqrt{a^2 + b^2} \sin (Bt + C) \). (b) Find the amplitude of the oscillations of the weight. (c) Find the frequency of the oscillations of the weight.

Short Answer

Expert verified
The model in the form \( y = \sqrt{a^2 + b^2} \sin (Bt + C) \) is \( y = \dfrac{5}{12} \sin (2t + \arctan (\dfrac{3}{4})) \). The amplitude of the oscillations of the weight is \( \dfrac{5}{12} \) feet and the frequency of the oscillations is 2 oscillations per second.

Step by step solution

01

Rewrite the given model

Use the identity \( a \sin B \theta + b \cos B \theta = \sqrt{a^2 + b^2} \sin (B\theta + C) \), where \( C = \arctan (b/a) \), \( a > 0 \), to rewrite the given model \( y = \dfrac{1}{3} \sin 2t + \dfrac{1}{4} \cos 2t \) in the form $ y = \sqrt{a^2 + b^2} \sin (Bt + C) $. In this case, \( a = \dfrac{1}{3} \), \( b = \dfrac{1}{4} \), \( B = 2 \). So \( C = \arctan (\dfrac{1}{4}/\dfrac{1}{3}) = \arctan (\dfrac{3}{4}) \). Thus the model becomes \( y = \sqrt{(\dfrac{1}{3})^2 + (\dfrac{1}{4})^2} \sin (2t + \arctan (\dfrac{3}{4}))
02

Find the amplitude of the oscillations

The amplitude of the oscillations of the weight can be found from the coefficient of the sine function in the rewritten model. Rewrite the square under the root as a fraction so it can be calculated. Compute \( \sqrt{(\dfrac{1}{3})^2 + (\dfrac{1}{4})^2} = \sqrt{\dfrac{1}{9} + \dfrac{1}{16}} = \sqrt{\dfrac{25}{144}} = \dfrac{5}{12} \). The amplitude is \( \dfrac{5}{12} \) feet.
03

Find the frequency of the oscillations

The frequency of the oscillations of the weight is represented by the coefficient of \( t \) in the argument of the sine function in the rewritten model. In this case, the frequency is given by the value of \( B \) which equals 2. The frequency is therefore 2 oscillations per second.

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Key Concepts

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

Trigonometric Identity
In precalculus, trigonometric identities are formulas that relate the trigonometric functions to one another. They allow us to simplify complex expressions, solve trigonometric equations, and model real-world phenomena. A commonly used identity, especially in the context of harmonic motion, is the sum-to-product identity.

This specific identity allows us to combine a sine and cosine term into a single sine term with an amplitude and a phase shift. For the given exercise, the sum-to-product identity is stated as:
\[ a \sin(B\theta) + b \cos(B\theta) = \sqrt{a^2 + b^2} \sin(B\theta + C) \]
Here, \(C\) is the phase shift calculated as \(C = \arctan(b/a)\) when \(a > 0\). This identity is extremely useful when modeling harmonic motion, as it allows us to express the motion's path in a clear and standardized form, enabling further analysis such as finding the amplitude and frequency of the oscillations.
Amplitude of Oscillations
The amplitude of oscillations is a fundamental concept in understanding harmonic motion. It represents the maximum extent of a vibration or oscillation, measured from the position of equilibrium. In a mathematical sense, it is the coefficient of the sine or cosine function in a harmonic equation.

The equation given in the exercise, \( y = \dfrac{1}{3} \sin 2t + \dfrac{1}{4} \cos 2t \), was restructured using the trigonometric identity to find the amplitude. The calculation involved finding the square root of the sum of the squares of the coefficients of the sine and cosine terms, thus yielding the amplitude:
\[ \sqrt{\left(\dfrac{1}{3}\right)^2 + \left(\dfrac{1}{4}\right)^2} = \sqrt{\dfrac{1}{9} + \dfrac{1}{16}} = \dfrac{5}{12} \]
The amplitude relates directly to the physical context of the problem—the higher the amplitude, the stronger the oscillations. In this scenario, the amplitude signifies that the weight on the spring moves up to \( \dfrac{5}{12} \) feet away from its equilibrium position.
Frequency of Oscillations
Frequency, often measured in cycles per second or Hertz (Hz), is defined as the number of oscillations that occur in a unit of time. In the context of harmonic motion, it is an essential measure that indicates how fast a weight oscillates back and forth from its equilibrium position.

With reference to the given exercise, the frequency of oscillations can be determined from the coefficient of \( t \) within the sine function. After applying the trigonometric identity to the original model, the frequency is represented by the term \( B \) in the argument of the sine function. For the model:
\( y = \sqrt{\left(\dfrac{1}{3}\right)^2 + \left(\dfrac{1}{4}\right)^2} \sin(2t + \arctan(\dfrac{3}{4})) \)
The coefficient \( 2 \) directly indicates the frequency of the motion. Thus, the calculated frequency is 2 cycles per second or 2 Hz. This means that the weight completes two full cycles of oscillation every second, which is a relatively high rate of oscillation and typical for systems such as springs and pendulums used in harmonic motions.

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