- #1
UrbanXrisis
- 1,196
- 1
the question is http://home.earthlink.net/~urban-xrisis/phy002.jpg
a.
[tex]\omega = \sqrt{\frac{k}{m}}[/tex]
[tex]\omega = \sqrt{\frac{500}{2}}[/tex]
b.
[tex]F=kx[/tex]
[tex]\frac{g}{3} * 2kg=500x[/tex]
[tex]x=1.31cm[/tex]
c.
since the block displaces 1.31 cm then that is it's amplitude
I don't undersatnd what they mean by inital phase angle. The initial phase would be the mass at -1.31cm moving upwards. I would just be...
[tex]x=- \omega^2 A cos(\omega t + \phi)[/tex]
[tex] \phi=0[/tex]
or... I think this might be a valid answer...
[tex]x_i=Acos\phi[/tex]
[tex]-1.31=1.31cos\phi[/tex]
[tex] \phi = \pi[/tex]
am I doing this all correctly?
a.
[tex]\omega = \sqrt{\frac{k}{m}}[/tex]
[tex]\omega = \sqrt{\frac{500}{2}}[/tex]
b.
[tex]F=kx[/tex]
[tex]\frac{g}{3} * 2kg=500x[/tex]
[tex]x=1.31cm[/tex]
c.
since the block displaces 1.31 cm then that is it's amplitude
I don't undersatnd what they mean by inital phase angle. The initial phase would be the mass at -1.31cm moving upwards. I would just be...
[tex]x=- \omega^2 A cos(\omega t + \phi)[/tex]
[tex] \phi=0[/tex]
or... I think this might be a valid answer...
[tex]x_i=Acos\phi[/tex]
[tex]-1.31=1.31cos\phi[/tex]
[tex] \phi = \pi[/tex]
am I doing this all correctly?
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