Oscillations and Physical Pendulum help

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SUMMARY

The discussion focuses on solving problems related to oscillations and physical pendulums, specifically using the equations of simple harmonic motion and the parallel-axis theorem. The first problem involves a piston described by the equation x = (7.00 cm) cos(5t + π/8), where participants clarify the correct approach to find position, velocity, and acceleration at t=0. The second problem addresses the period of oscillation for a rigid rod using the formula T = 2π√(I/mgd), with participants discussing the correct moment of inertia and distance from the center of mass.

PREREQUISITES
  • Understanding of simple harmonic motion equations
  • Familiarity with the parallel-axis theorem
  • Knowledge of moment of inertia calculations
  • Ability to use trigonometric functions in radians
NEXT STEPS
  • Study the derivation of the simple harmonic motion equations
  • Learn about the parallel-axis theorem in detail
  • Explore different methods for calculating moment of inertia
  • Practice converting between degrees and radians in trigonometric calculations
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Students in physics, educators teaching mechanics, and anyone interested in understanding oscillatory motion and pendulum dynamics.

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1) SOLVED. In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression, where x is in centimeters and t is in seconds. x = (7.00 cm) cos(5t + π/8). (a) at t=0, find the position (b) velocity (c) acceleration

For part a, I tried plugging in 0 for t, giving 7cos(0 + pi/8), which equals 6.999 or 7? That's the only thing I know to do, and this is not the correct answer

For b and c, I tried plugging in 0 to the equations v = -wAsin(wt + [tex]\phi[/tex]) and a = -w2Acos(wt + [tex]\phi[/tex]), but once again no luck.

2) A very light rigid rod with a length of 1.81 m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation. Ip = ICM + MD2 (a) Determine the period of oscillation. [Suggestion: Use the parallel-axis theorem equation given. Where D is the distance from the center-of-mass axis to the parallel axis and M is the total mass of the object.]

[tex]T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{I}{mgd}}[/tex] is what I know I need to be using, but I'm not sure what to plug in for I. I tried plugging in [tex]\frac{1}{3}ML^{2}[/tex] for I, giving me [tex]2\pi\sqrt{\frac{\frac{1}{3}ML^{2}}{mgd}}[/tex]
The mass canceled out and I changed d to L, which canceled out and reduced L2 to L. When I plugged in the numbers I didn't get the right answer. Is my value for d wrong or is it the I as a whole?

Any help is appreciated.
 
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kbump said:
1) In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression, where x is in centimeters and t is in seconds. x = (7.00 cm) cos(5t + π/8). (a) at t=0, find the position (b) velocity (c) acceleration

For part a, I tried plugging in 0 for t, giving 7cos(0 + pi/8), which equals 6.999 or 7? That's the only thing I know to do, and this is not the correct answer
Check to see if your calculator is in radian or degree mode! pi/8= .392... degrees would be very close to 0 so cos(pi/8) would be very close to 1. cosine of pi/8 radians however, is not 1 so 7 cos(pi/8) is not 7.

For b and c, I tried plugging in 0 to the equations v = -wAsin(wt + [tex]\phi[/tex]) and a = -w2Acos(wt + [tex]\phi[/tex]), but once again no luck.
Again, put your calculator into radian mode.

2) A very light rigid rod with a length of 1.81 m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation. Ip = ICM + MD2 (a) Determine the period of oscillation. [Suggestion: Use the parallel-axis theorem equation given. Where D is the distance from the center-of-mass axis to the parallel axis and M is the total mass of the object.]

[tex]T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{I}{mgd}}[/tex] is what I know I need to be using, but I'm not sure what to plug in for I. I tried plugging in [tex]\frac{1}{3}ML^{2}[/tex] for I, giving me [tex]2\pi\sqrt{\frac{\frac{1}{3}ML^{2}}{mgd}}[/tex]
The mass canceled out and I changed d to L, which canceled out and reduced L2 to L. When I plugged in the numbers I didn't get the right answer. Is my value for d wrong or is it the I as a whole?

Any help is appreciated.
 
You were right I had my calculator in degrees, thanks! #1 is down, now all I need help on is #2!
 

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