Simple harmonic motion of a machine part

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Homework Help Overview

The discussion revolves around a machine part undergoing simple harmonic motion (SHM) with a specified frequency and amplitude. The original poster seeks to determine the time it takes for the part to move from a position of x = 0 to x = -1.80 cm.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between displacement and amplitude, with one noting that the displacement of -1.8 cm corresponds to the negative amplitude. Others suggest sketching a displacement vs. time graph to visualize the motion and identify the timing of specific displacements.

Discussion Status

Several participants have provided insights into the problem, including the identification of the period of motion and the suggestion that there are multiple valid answers for the time taken to reach the specified displacement. The conversation reflects a productive exploration of the topic without reaching a definitive conclusion.

Contextual Notes

Participants are working within the constraints of the problem as posed, discussing the implications of the amplitude and the periodic nature of SHM. There is an acknowledgment of the potential for multiple valid answers based on the nature of the motion.

kaywond
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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?


Homework Equations


x = Acoswt


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?
 
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The displacement is the negative of the amplitude.
 
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?
 
thanks 1/5 equals 0.2. Then divide by 4 and 0.05 is the time in seconds
 
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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