### Determining the Velocity from a Position FunctionIn physics, if the position \( x \) of a particle undergoing simple harmonic motion is given by:\[ x = A \cos(\omega t + \phi) \]where:- \( A \) is the amplitude,- \( \omega \) is the angular frequency,- \( t \) is the time, and- \( \phi \) is the phase constant,then the velocity \( v \) can be determined by differentiating the position function with respect to time \( t \).Here are the given options to determine \( v \):a. \( -\omega A \sin(\omega t + \phi) \)b. \( \omega A \cos(\omega t + \phi) \)c. \( \omega^2 A \sin(\omega t + \phi) \)d. \( +\omega A \sin(\omega t + \phi) \)e. \( -\omega^2 A \cos(\omega t + \phi) \)### Explanation:To find the velocity \( v \), we differentiate the given position function \( x \) with respect to time \( t \):\[ v = \frac{dx}{dt} \]Given:\[ x = A \cos(\omega t + \phi) \]Using the chain rule, we have:\[ v = A \cdot \frac{d}{dt} \left[ \cos(\omega t + \phi) \right] \]Since the derivative of \( \cos(\omega t + \phi) \) with respect to \( t \) is:\[ \frac{d}{dt} \left[ \cos(\omega t + \phi) \right] = -\omega \sin(\omega t + \phi) \]Therefore:\[ v = A \cdot (-\omega \sin(\omega t + \phi)) \]\[ v = -\omega A \sin(\omega t + \phi) \]### Correct Answer:a. \( -\omega A \sin(\omega t + \phi) \)

Answered: If x= Acos(at +6) then v is a. -aA…

If x= Acos(at +6) then v is a. -aA sin(at +ø). b. aA cos(at + 0). c. w²A sin(at +9). d. +aA sin(at + ø). e. -w*A cos(at +ø).

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Question

### Determining the Velocity from a Position Function

In physics, if the position \( x \) of a particle undergoing simple harmonic motion is given by:

\[ x = A \cos(\omega t + \phi) \]

where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is the time, and
- \( \phi \) is the phase constant,

then the velocity \( v \) can be determined by differentiating the position function with respect to time \( t \).

Here are the given options to determine \( v \):

a. \( -\omega A \sin(\omega t + \phi) \)
b. \( \omega A \cos(\omega t + \phi) \)
c. \( \omega^2 A \sin(\omega t + \phi) \)
d. \( +\omega A \sin(\omega t + \phi) \)
e. \( -\omega^2 A \cos(\omega t + \phi) \)

### Explanation:
To find the velocity \( v \), we differentiate the given position function \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]

Given:
\[ x = A \cos(\omega t + \phi) \]

Using the chain rule, we have:
\[ v = A \cdot \frac{d}{dt} \left[ \cos(\omega t + \phi) \right] \]

Since the derivative of \( \cos(\omega t + \phi) \) with respect to \( t \) is:
\[ \frac{d}{dt} \left[ \cos(\omega t + \phi) \right] = -\omega \sin(\omega t + \phi) \]

Therefore:
\[ v = A \cdot (-\omega \sin(\omega t + \phi)) \]
\[ v = -\omega A \sin(\omega t + \phi) \]

### Correct Answer:
a. \( -\omega A \sin(\omega t + \phi) \)

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