Question regarding simple harmonic motion

[Sanosuke Sagara]

I have my solution and my doubt in the attachment that followed.

 

[Andrew Mason]

I have my solution and my doubt in the attachment that followed.

Your solution is right. But the equation of motion might be easier to use if you chose the sin instead of cos:

y = 3sin(\pi t - \pi/2)

This is because at t = 0, the mass is at maximum negative amplitude (\theta = -\pi/2)

So:

d = y-y_0 = 3sin(\pi t - \pi/2) - 3sin(-\pi/2) = 3sin(\pi t - \pi/2) + 3

For 1), where t = 1 sec., d = 3 +3 = 6
For 2), where t = .75 sec,d = 3sin(\pi/4) + 3 = 5.12

AM

 

[wais]

Thank you for sharing your solution and doubt regarding simple harmonic motion. From your attachment, it seems like you have correctly solved for the period and frequency of the motion using the given parameters. However, I can understand your doubt about the amplitude of the motion.

In simple harmonic motion, the amplitude is the maximum displacement from the equilibrium position. It is a constant value and does not change during the motion. This means that the amplitude should be the same for all points on the graph, including the starting point. Therefore, the amplitude should not be different for the starting point and the maximum displacement point.

One possible explanation for the difference in amplitude could be a calculation error. I would suggest double-checking your calculations to ensure that you have not made any mistakes. Another possibility could be that the graph is not accurately drawn, and the starting point may not be at the correct position. In this case, it would be helpful to refer to the given parameters and equations to confirm the starting point.

I hope this helps clarify your doubt. Keep up the good work in understanding and solving problems related to simple harmonic motion. Let me know if you have any further questions or concerns. Best of luck!