Time it takes an oscillating mass to reach a certain velocity

AI Thread Summary
The discussion revolves around solving for the time at which an object in simple harmonic motion reaches a velocity of -0.25 m/s, given the velocity function vx(t) = -0.35sin(20t + π). Initial attempts to solve the equation resulted in a negative time, which was incorrect. Participants emphasized the periodic nature of the sine function and the importance of correctly applying the arcsine and adjusting for the phase shift. Ultimately, the correct time was approximated to be around 0.2 seconds after accounting for necessary adjustments in calculations. The participants encouraged a deeper understanding of trigonometric functions to resolve similar issues in the future.
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Hi everybody, this problem is giving me trouble and I was wondering if you could give me any advice.

Homework Statement



The velocity of an object in simple harmonic motion is given by vx(t)= - (0.35 m/s)sin(20t + pi), where t is in s.

What is the first time after t=0s at which the velocity is - 0.25 m/s?

Homework Equations



See the equation above.

The Attempt at a Solution



I simply plugged -.25 m/s in for vx(t) above and solved for t, so I got:

-0.25 = -0.35sin(20t + pi)
.714 = sin(20t + pi)

When I solve for t I get t = -.397e-1 = -.0397s, but mastering physics (the program I use to do homework online) say it's wrong. I entered the answer as positive. Anybody have any idea? Thanks a lot!

edit: I just went to see my professor and he helped me set it up, when finished we had sin(20t) = -5/7, but master physics says it's wrong.
 
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Trigonometric functions are periodic. This means that the function "repeats" itself over and over as you move along the X axis. If you visualize the sine involved in your problem, you'll realize that if you move sufficiently far to the right from -.0397 (and beyond 0), you'll get at a point p where vt(p) = vt(-.0397). That point is in fact -.0397 + Pi/10, because Pi/10 is the period of vt(x).
 
Thanks for the reply. Does that mean that the time it reaches -.25m/s is .275? (-0.397 + pi/10). Where did that 10 come from? Thanks!
 
The period of a sine function is given by 2pi/k, where k is the coefficient of x inside the bracket (i.e. for vt(x), the coefficient is 20). Really, it's not very useful for you to hear about this from me. You'd be better off learning better about trigonometric functions.
 
I believe that I do have a decent understand of trigonometric functions, I just can't figure out why this problem is giving me so much trouble. I can't be doing much wrong...]

I've got a graph of the wave drawn out and where the time will fall on the wave, but I just can't get it numerically.
 
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But you just did; .275 s.
 
Thanks for the reply, but this darn program say it's wrong. I really don't like doing homework online.
 
Did you calculate arcsine in radians?
 
Yeah, arcsine(5/7) is .7956, then multiplying that my (1/20) as by my answer in the first post gives .0397
 
  • #10
Yeah, arcsine(5/7) is .7956, then multiplying that my (1/20) as by my answer in the first post gives .0397

You forgot to subtract pi before dividing by 20.

Calculating it myself, I get around -0.12 for t. -0.12 + pi/10 = 0.2, approximately.
 
  • #11
Thanks a lot for the help, Werg. Any more input from anybody would be appreciated!
 
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