How is the Answer to 3(d) Found in Simple Harmonic Motion Problem?

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Homework Help Overview

The discussion revolves around a simple harmonic motion problem, specifically focusing on how to determine the answer to part 3(d) of the question. Participants are exploring the calculation of velocity from a position versus time function, with a particular emphasis on discrepancies between their calculations and the provided answer.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating velocity using the gradient of a tangent to the curve and derivatives. Questions arise regarding the validity of their calculations and the correctness of the answer key, particularly in relation to the amplitude and period of the motion.

Discussion Status

The discussion is active, with participants sharing their calculations and questioning the accuracy of the provided answer. Some participants suggest that the answer in the book may be incorrect, while others seek clarification on the methods used to derive velocity.

Contextual Notes

Participants mention that certain equations, such as v = -Aw sin(wt), are not included in the chapter of the exercise, which may affect their ability to solve the problem. There is also a note that some participants are studying independently without access to instructors or teaching assistants.

tahmidbro
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Homework Statement
Simple harmonic motion problem:
How is the answer to 3 (d) is found?
Relevant Equations
N/A
How is the answer to 3 (d) is found?
 

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tahmidbro said:
Homework Statement:: Simple harmonic motion problem:
How is the answer to 3 (d) is found?
Relevant Equations:: N/A

How is the answer to 3 (d) is found?
Welcome to PF.

Are you familiar with how to find velocity from a position versus time function?
 
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Yes. I have drawn a tangent with the curve at 'z' and calculated the gradient.
Gradient = ( 3+3 )/(7-5) = 3 cm/s = 0.03m/s.
But the answer is 6m/s. Will you please tell me how?
 
tahmidbro said:
Yes. I have drawn a tangent with the curve at 'z' and calculated the gradient.
Gradient = ( 3+3 )/(7-5) = 3 cm/s = 0.03m/s.
But the answer is 6m/s. Will you please tell me how?
I have used a derivative to calculate the velocity at that point, and also get a much smaller number than the answer key. If the amplitude is 3cm and the period is 8 seconds, the peak velocity is much less than the 6m/s that is listed as the answer.

Do you know how to do this calculation with a derivative? Also, can you check that answer with the instructor or a teaching assistant?
 
Will you please share your calculation here? are you using v = -Aw sin(wt) ? ( this equation is not in the chapter of the exercise )
well, I am studying by myself. I do not have any instructor or a teaching assistant.
 
tahmidbro said:
Will you please share your calculation here? are you using v = -Aw sin(wt) ? ( this equation is not in the chapter of the exercise )
Yes, good. Differentiating that cos() function does give that derivative.
 
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w = 2pi/8 =0.785
A = 3/100 metre . For maximum velocity at 'z', sinwt = 1
v= Aw = 0.785 x 3/100 = 0.0235 m/s
But the answer from the book is 6 m/s. How do I get that?
 
As I said, it looks like the answer provided is wrong. And not by a little bit!
 
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berkeman said:
As I said, it looks like the answer provided is wrong. And not by a little bit!
Okay , Thanks for the help!
 
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