- #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?

Last edited by a moderator: