College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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I am lost on how to find the equation to derive for part c of this problem. Though the solution a solution is offered (to take dv/dt by substituting t=0 into VI) no equation is given for eq. VI. I'm unable to work backwards from -17.3cm/s², and would appreciate some guidance.
![**Chapter 15: Problem 4**
**Problem Statement:**
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression:
\[ x = 5.00 \cos \left(2t + \frac{\pi}{6}\right) \]
This can be simplified to the form:
\[ x = A \cos(\omega t + \phi) \]
where \( x \) is in centimeters and \( t \) is in seconds. At \( t = 0 \), find (a) the position of the particle, (b) its velocity, and (c) its acceleration. Find (d) the period and (e) the amplitude of the motion.
---
**Solution:**
4. Given the standard form \( A \cos(\omega t + \phi) \):
a) At \( t = 0 \):
\[ x = 5 \cos(2(0) + \frac{\pi}{6}) = 5 \cos \frac{\pi}{6} \approx 4.999 \, \text{cm} = X \]
b) To find velocity \( V \):
\[ V = \frac{dx}{dt} \]
Using:
\[ V = x' = \frac{d}{dt} [A \cos(2t + \phi)] = 2A[-\sin(2t + \phi)] \]
\[ = -2 \times 5 \sin(2t + \frac{\pi}{6}) \Rightarrow -10 \sin(\frac{\pi}{6}) \]
\[ = -10(\frac{1}{2}) = -5 \, \text{cm/s} = V \]
c) For acceleration \( a \):
\[ a = \frac{dv}{dt} \]
d) To find the period \( T \):
\[ T = \frac{2\pi}{\omega} \Rightarrow \omega = \frac{2\pi}{T} \]
From the given equation:
\[ \omega = 2 \]
\[ T = \frac{2\pi}{2} = \pi \approx 3.14 \, \text{s} = T \]
e) Amplitude \( A \) is given as:
\[ A = 5 \, \text{cm} \]
---
In this problem, we are analyzing the](https://content.bartleby.com/qna-images/question/c4fa76a1-6d04-4636-9f2a-e2a5fe8d46c0/9a1f693b-a73d-4574-93a8-dca3722c7ec5/3qi4vr_processed.jpeg)
Transcribed Image Text:**Chapter 15: Problem 4**
**Problem Statement:**
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression:
\[ x = 5.00 \cos \left(2t + \frac{\pi}{6}\right) \]
This can be simplified to the form:
\[ x = A \cos(\omega t + \phi) \]
where \( x \) is in centimeters and \( t \) is in seconds. At \( t = 0 \), find (a) the position of the particle, (b) its velocity, and (c) its acceleration. Find (d) the period and (e) the amplitude of the motion.
---
**Solution:**
4. Given the standard form \( A \cos(\omega t + \phi) \):
a) At \( t = 0 \):
\[ x = 5 \cos(2(0) + \frac{\pi}{6}) = 5 \cos \frac{\pi}{6} \approx 4.999 \, \text{cm} = X \]
b) To find velocity \( V \):
\[ V = \frac{dx}{dt} \]
Using:
\[ V = x' = \frac{d}{dt} [A \cos(2t + \phi)] = 2A[-\sin(2t + \phi)] \]
\[ = -2 \times 5 \sin(2t + \frac{\pi}{6}) \Rightarrow -10 \sin(\frac{\pi}{6}) \]
\[ = -10(\frac{1}{2}) = -5 \, \text{cm/s} = V \]
c) For acceleration \( a \):
\[ a = \frac{dv}{dt} \]
d) To find the period \( T \):
\[ T = \frac{2\pi}{\omega} \Rightarrow \omega = \frac{2\pi}{T} \]
From the given equation:
\[ \omega = 2 \]
\[ T = \frac{2\pi}{2} = \pi \approx 3.14 \, \text{s} = T \]
e) Amplitude \( A \) is given as:
\[ A = 5 \, \text{cm} \]
---
In this problem, we are analyzing the
![**Expert Solution**
**(c)**
**To determine**
The acceleration of the piston.
**Answer to Problem 14P**
The acceleration of the piston is \(-17.3 \, \text{cm/s}^2\).
**Explanation of Solution**
Write the expression for the acceleration.
\[ a = \frac{dv}{dt} \quad (V) \]
Use equation (II) in equation (V),
\[ a = \quad (VI) \]
**Conclusion:**
Substitute \( t = 0 \) in equation (VI),
\[ a = -17.3 \, \text{m/s}^2 \]
Therefore, the acceleration of the piston is \(-17.3 \, \text{cm/s}^2\).
*Note: The image does not contain any graphs or diagrams.*](https://content.bartleby.com/qna-images/question/c4fa76a1-6d04-4636-9f2a-e2a5fe8d46c0/9a1f693b-a73d-4574-93a8-dca3722c7ec5/a2tpx4e_processed.jpeg)
Transcribed Image Text:**Expert Solution**
**(c)**
**To determine**
The acceleration of the piston.
**Answer to Problem 14P**
The acceleration of the piston is \(-17.3 \, \text{cm/s}^2\).
**Explanation of Solution**
Write the expression for the acceleration.
\[ a = \frac{dv}{dt} \quad (V) \]
Use equation (II) in equation (V),
\[ a = \quad (VI) \]
**Conclusion:**
Substitute \( t = 0 \) in equation (VI),
\[ a = -17.3 \, \text{m/s}^2 \]
Therefore, the acceleration of the piston is \(-17.3 \, \text{cm/s}^2\).
*Note: The image does not contain any graphs or diagrams.*
Expert Solution
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Step 1
Position for simple harmonic function is given
x=5.00cos(2t+π6) .....(a)
(b)
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