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- Thread starter tahmidbro
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In summary, the answer to 3 (d) is found by calculating the gradient of the position versus time function, and differentiating that cos() function.

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Welcome to PF.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?

Are you familiar with how to find velocity from a position versus time function?

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Gradient = ( 3+3 )/(7-5) = 3 cm/s = 0.03m/s.

But the answer is 6m/s. Will you please tell me how?

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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.tahmidbro said:

Gradient = ( 3+3 )/(7-5) = 3 cm/s = 0.03m/s.

But the answer is 6m/s. Will you please tell me how?

Do you know how to do this calculation with a derivative? Also, can you check that answer with the instructor or a teaching assistant?

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well, I am studying by myself. I do not have any instructor or a teaching assistant.

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Yes, good. Differentiating that cos() function does give that derivative.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 )

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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?

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As I said, it looks like the answer provided is wrong. And not by a little bit!

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Okay , Thanks for the help!berkeman said:As I said, it looks like the answer provided is wrong. And not by a little bit!

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