how to find the equation to derive for part c of this problem

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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
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.*
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
Step 1

Position for simple harmonic function is given

x=5.00cos2t+π6      .....(a)

(b) v=dxdt

  =5-1×2×sin2t+π6=-10sin2t+π6     

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