What is the Period of Oscillation Based on Given Coordinates?

AI Thread Summary
The discussion focuses on calculating the period of oscillation based on given coordinates for points R and K. The x-coordinate of point R is 0.12m, and the time coordinate for point K is 0.0050s. It is determined that the time from R to K represents a quarter of the total period, leading to the calculation of the total period as 0.02s. The conversation emphasizes understanding the relationship between distance, time, and the sinusoidal waveform's properties. The final question highlights the significance of this calculated period in the context of oscillatory motion.
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Homework Statement



Now assume that the x coordinate of point R is 0.12m and the t coordinate of point K is 0.0050 s.

What is the period T ?

Homework Equations



T = 1/f where f is the frequency and T is the period.

The Attempt at a Solution



From the graph, there are 2 periods in oscillation. So I guess that the period would be 2T and since it starts at R = .12m and reaches point K in .0050 s . So the velocity is .12m/.0050 s = 24m/s

Not sure if this is correct but, T = 2T = .12m/.005s = T = 12 s
 

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No. The horizontal distance from 0 to K is a quarter of the period of that sinusoidal waveform. Since you are given the time difference from 0 to K, and that is a quarter of the period, what is the total period?
 
berkeman said:
No. The horizontal distance from 0 to K is a quarter of the period of that sinusoidal waveform. Since you are given the time difference from 0 to K, and that is a quarter of the period, what is the total period?

Thank for the help

From what you have said, I can calculate the total period to be (.005)4 = .02s
 
Good. Quiz Question -- what is special about that period?
 

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