# Simple harmonic motion problem

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

#### tahmidbro

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?

#### Attachments

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?

• etotheipi
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.

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.
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.

• tahmidbro
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!

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

• berkeman