Solve Pendulum Problem: Find Position in 1.6 Sec

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Homework Help Overview

The discussion revolves around a simple harmonic motion (SHM) problem involving a pendulum released from a 15-degree angle and its position after 1.6 seconds, with a frequency of 2 Hz. Participants are exploring the relationship between angular displacement and arc length in the context of pendulum motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of amplitude and the use of the equation X = A cos(2πft) to find the position. There are questions about the interpretation of amplitude in degrees versus meters, and the implications of using linear approximations for larger angles. Some participants suggest alternative methods or express concerns about the complexity of the problem.

Discussion Status

The discussion is active, with participants providing various insights and questioning the assumptions made in the problem setup. Some guidance has been offered regarding the use of the cosine function and the relationship between angle and arc length, but no consensus has been reached on the correct approach or interpretation.

Contextual Notes

There are indications of confusion regarding the dimensions of amplitude and the appropriateness of using linear approximations for larger angles in pendulum motion. The original poster's calculations and the textbook's expected results are also under scrutiny.

beanryu
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I got a SHM problem:

a pendulum is released 15 degrees from the vertical and has a frequency of 2, what is its position in 1.6 sec. Hint: (don't confuse the swing angle with the argument of cosine)

I found that the amplitude(A) is an arc of about 0.017m.

I used the equation: X = A cos ( 2 "pie" f t )
but my answer is wrong. I got X=0.016 and it represent an angle of about 0.88 degrees. But according to the book, it should be 4.6 degrees away from the equilibrium point.

In addition, can someone tell me what does the "argument of cosine" mean?

Thank you for replying!
 
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the amplitude(A) is in dimension of degree, not meter... tell me where is your .0017 m came from ? ...
 
The problem is wrongly formulated...For an amplitude of oscillation if 15°,the linear approximation will not hold...In this case the solution is expressible through Jacobi's elliptic functions...

Daniel.
 
T = 1/f

T = 2 "pie" square root of (L/g)

I got L is about 0.06 meter, which is the length of the string where the weight is hanging.
I calculated the circumference using C = 2 "pie" r, where I used the value of L as the value of the radius, and got the result of about 0.38 meter.
Then I used the following ration : (x/0.38)=(15degrees/360degrees) and found x to be an arc that covers the amplitude in about 0.017 meter.

I think my way is kind of long and maybe there's another short way to finger it out, but this is how I did it, and can anyone provide further suggestion or advices?

Maybe you are right, Dextercioby, the textbook said its about 15 degrees.

Thank you
 
the exact solution for a large angle is not easy.. and not for your level... forget it...
since the arc length is directly proportional to the angle (hope you can see that). It is meaningless to find the arclength. you equation X=Acos(2pi f t) works as well as angle or the arc length. put A = 15 degree, X will be your answer...
 
okay, thank you vincentchan.
 

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