- #1
cj
- 85
- 0
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?
[tex]\omega = \omega_0[/tex]
has maximum amplitude U and positive value.
Find x(t).
----------
The textbook answer for this problem is:
[tex]x(t) = -\frac {U}{\omega_0}sin(\omega_0t)[/tex]
where
[tex]U = \omega_0A}[/tex]
----------
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?