Solve Pendulum Problem: Find Position in 1.6 Sec

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The discussion revolves around solving a simple harmonic motion (SHM) problem involving a pendulum released at a 15-degree angle with a frequency of 2. The user initially calculated the position using the equation X = A cos(2πft) but arrived at an incorrect answer, prompting confusion about the amplitude's unit and the meaning of "argument of cosine." Clarifications indicate that the amplitude should be treated in degrees rather than meters, and the solution for large angles requires more complex functions than simple linear approximations. Participants suggest that the equation can be used directly with the amplitude set to 15 degrees for a more straightforward solution. The conversation highlights the challenges of accurately modeling pendulum motion at larger angles.
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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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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