Position of an oscillating object

  • Thread starter ChloeYip
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  • #1
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Homework Statement



The position of an object that is oscillating on an ideal spring is given by the equation x =
(12.3 cm) cos[(1.26s-1)t]. At time t = 0.815 s,
(a) how fast is the object moving?
(b) what is the magnitude of the acceleration of the object?

Homework Equations


As follow

The Attempt at a Solution


dx/dt = -12.3*(1.26)sin(1.26*.815) = -0.2777 cm/s but the answer is -13.3cm/sec
dx^2/dt^2= -12.3*(1.26)^2*cos(1.26*.815) = -19.524 cm/s^2 but the answer is -10.1cm/sec^2

I have only little time before test.hope there is someone help me soon.tell me whats wrong with it.
(Please don't ask me to guess... i m not good at that and i really have not time left...)
Thank you very much for helping me
 

Answers and Replies

  • #2
vela
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##12.3 \ne 1.23## for one thing.
 
  • #3
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Thanks for reminding
Thats only multiple of 10, still canet get the answer
Also, the calculated answer is not affect by this typo
Sorry for typo
 
  • #4
vela
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Make sure your calculator is in radian mode.
 
  • #5
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Ahhh yes
Thanksssssss
 

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