/*! 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 40 The displacement of a particle i... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 40

The displacement of a particle is represented by the equation \(y=3 \cos \left(\frac{\pi}{2}-2 \omega t\right)\). The motion of the particle is (A) simple harmonic with period \(2 \pi / \omega\). (B) simple harmonic with period \(\pi / \omega\). (C) periodic but not simple harmonic. (D) non-periodic.

Short Answer

Expert verified
The motion of the particle is simple harmonic with a period of \(\frac{\pi}{\omega}\) (option B).

Step by step solution

01

Identify the given displacement equation

Displacement is given by the equation: \( y=3 \cos\left(\frac{\pi}{2}-2 \omega t\right) \)
02

Recognize the equation's form

Comparing the equation to the general form of simple harmonic motion (SHM): \( y(t) = A \cos(\omega t + \phi) \) or \( y(t) = A \sin(\omega t + \phi) \) Where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase angle. The given displacement equation is in the form of the cos function, which implies it can represent simple harmonic motion.
03

Rewrite the equation

Next, we will rewrite the given equation in a simplified form to make it easier to compare with SHM equations: \( y = 3 \cos\left(\frac{\pi}{2} - 2\omega t \right) \) We can rewrite this using the trigonometric identity \( \cos(x - y) = \cos(x)\cos(y) + \sin(x)\sin(y) \): \( y = 3\left[\cos\left(\frac{\pi}{2}\right)\cos(2\omega t) + \sin\left(\frac{\pi}{2}\right)\sin(2\omega t)\right] \) Since \(\cos(\frac{\pi}{2}) = 0\) and \(\sin(\frac{\pi}{2}) = 1\), the equation simplifies to: \( y = 3\sin(2\omega t) \)
04

Compare to simple harmonic motion

Now, we compare the simplified equation to the general equation of simple harmonic motion: \( y = 3\sin(2\omega t) \) 鈫� \( y(t) = A \sin(\omega t + \phi ) \) In our equation, \(A = 3\) which is the amplitude, and \(\omega' = 2\omega\) which represents the angular frequency of the given equation.
05

Determine the period and answer

The period of simple harmonic motion can be found using the following equation: \( T = \frac{2\pi}{\omega'} \) For our given equation, the angular frequency is \(\omega' = 2\omega\), so the period becomes: \( T = \frac{2\pi}{2\omega} \) \( T = \frac{\pi}{\omega} \) This matches option (B), which suggests the particle has simple harmonic motion with a period of \(\frac{\pi}{\omega}\). Hence, the answer is: (B) simple harmonic with a period of \(\frac{\pi}{\omega}\).

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