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[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?