- #1
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!
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!