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

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The discussion revolves around a simple harmonic motion problem where the calculated velocity at a specific point does not match the answer key. The participant calculated the gradient of the tangent to find the velocity but obtained 0.03 m/s instead of the expected 6 m/s. They used the derivative of the position function but still arrived at a much lower value. The conversation suggests that the answer provided in the book may be incorrect, as the calculations based on the given amplitude and period do not support it. The participant expresses gratitude for the assistance received in clarifying the calculations.
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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