- #1
erik-the-red
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Question:
An object is undergoing simple harmonic motion with period 0.305 s and amplitude 6.05 cm. At [tex]t=0[/tex] the object is instantaneously at rest at [tex]x=6.05[/tex] cm.
A.
Calculate the time it takes the object to go from [tex]x=6.05[/tex] cm to x = -1.49 cm.
I'm thinking [tex]\omega = \frac{2\pi}{T} = 20.6[/tex] rad/s. If we assume left to be the negative direction, then the [tex]\Delta x = -0.0754[/tex] m.
So, I can use the equation [tex]x = A\cos(\omega t + \phi)[/tex]. [tex]\phi = 0[/tex].
[tex]-0.0754 = .0605cos(20.6t)[/tex].
But, wait a minute here, this can't possibly be right. If I divide both sides by .0605, then I have cos equaling a magnitude greater than one.
I'm stuck.
An object is undergoing simple harmonic motion with period 0.305 s and amplitude 6.05 cm. At [tex]t=0[/tex] the object is instantaneously at rest at [tex]x=6.05[/tex] cm.
A.
Calculate the time it takes the object to go from [tex]x=6.05[/tex] cm to x = -1.49 cm.
I'm thinking [tex]\omega = \frac{2\pi}{T} = 20.6[/tex] rad/s. If we assume left to be the negative direction, then the [tex]\Delta x = -0.0754[/tex] m.
So, I can use the equation [tex]x = A\cos(\omega t + \phi)[/tex]. [tex]\phi = 0[/tex].
[tex]-0.0754 = .0605cos(20.6t)[/tex].
But, wait a minute here, this can't possibly be right. If I divide both sides by .0605, then I have cos equaling a magnitude greater than one.
I'm stuck.
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