Simple Oscillator: find x(t)

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  • #1
cj
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At time t = T/2 the speed of a simple oscillator of angular frequency

[tex]\omega = \omega_0[/tex]

has maximum amplitude U and positive value.

Find x(t).

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The textbook answer for this problem is:

[tex]x(t) = -\frac {U}{\omega_0}sin(\omega_0t)[/tex]

where

[tex]U = \omega_0A}[/tex]
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This is confounding me.

For a simple sin solution (I'm assuming zero phase shift), wouldn't x at t = T/4 have to be positive, i.e.:

[tex]x = \frac {U}{\omega_0}[/tex]

and NOT negative (as given by the alleged solution):

[tex]x = -\frac {U}{\omega_0}[/tex]

Comments?
 

Answers and Replies

  • #2
Integral
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What is the definition of the coordinate system?
 
  • #3
cj
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Vertical axis = displacement (x), velocity (v) and acceleration (a).

Horizontal axis = time, t
 
  • #4
Integral
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I am trying to get you to think about the definition of what is meant by negative and positive. Have you given us the complete problem?
 
  • #5
cj
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Integral said:
I am trying to get you to think about the definition of what is meant by negative and positive. Have you given us the complete problem?
Yes, the posted question and textbook answer (which I guess could be wrong) is verbatim.

Maybe I don't understand the meaning of the negative sign.

I'm imagining a displacement-speed-acceleration relationship as given in the graph at the top of the following linked page:

Sample

The bottom line is that if x(t) is as given by the textbook, it would clearly have a negative value (relative to the equilibrium position at t = T/4. This seems incorrect since even junior high school students understand that at t = T/4 the value for a sine function is clearly a maximum and clearly positive.

Right?
 

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