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Simple harmonic motion of a machine part

  • Thread starter kaywond
  • Start date
  • #1
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1. Homework Statement
A machine part is undergoing SHM with a frequency of of 5hz and amplitude of 1.80cm. How long does it take the part to go from x = 0 to -1.80cm?


2. Homework Equations
x = Acoswt


3. The Attempt at a Solution
X is given and convert it to metres 0.018. I need to manipulate answer to get time.
angular velocity is 2pi x 5 = 31.41. It is just an equation manipulation. could someone please show me the technique?
 

Answers and Replies

  • #2
The displacement is the negative of the amplitude.
 
  • #3
CWatters
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If the peak amplitude is 1.8cm then asking when the displacement is -1.8 makes it rather easy. No need to even use the sin button on the calculator :-)

Sketch displacement vs time. Mark the points 1.8m and -1.8m. If T is the period (=1/5) at what fraction of T does 1.8cm and -1.8cm occur?
 
  • #4
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thanks 1/5 equals 0.2. Then divide by 4 and 0.05 is the time in seconds
 
  • #5
CWatters
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0.05s is one valid answer but there are two possible valid right answers for this question. I recommend you give both.
 

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