Oscillations and Physical Pendulum help

In summary, the conversation involves two problems related to simple harmonic motion and the period of oscillation. In the first problem, the position, velocity, and acceleration of a piston in an engine are being analyzed. In the second problem, the period of oscillation of a meter stick suspended from a pivot is being determined using the parallel-axis theorem equation. The conversation includes troubleshooting for incorrect answers due to the use of degrees instead of radians in calculations.
  • #1
kbump
6
0
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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  • #2
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.
 
  • #3
You were right I had my calculator in degrees, thanks! #1 is down, now all I need help on is #2!
 

Related to Oscillations and Physical Pendulum help

1. What is an oscillation?

An oscillation is a repetitive back-and-forth motion around a central equilibrium point. It can occur in various forms, such as a swinging pendulum, a vibrating spring, or a bouncing ball.

2. How do you calculate the period of an oscillation?

The period of an oscillation is the time it takes for one complete cycle of motion. It can be calculated using the formula T = 2π√(l/g), where T is the period, l is the length of the pendulum or spring, and g is the acceleration due to gravity.

3. What is a physical pendulum?

A physical pendulum is a rigid body that is free to rotate about a fixed axis. Unlike a simple pendulum, which has a point mass at the end of a string, a physical pendulum has a mass distribution that affects its period and other characteristics.

4. How is the period of a physical pendulum affected by its mass distribution?

The period of a physical pendulum is affected by its mass distribution in two ways. Firstly, a greater mass distribution will increase the moment of inertia, which will result in a longer period. Secondly, the location of the center of mass affects the distance from the axis of rotation, which can also impact the period.

5. What factors can affect the accuracy of a physical pendulum experiment?

Some factors that can affect the accuracy of a physical pendulum experiment include air resistance, friction at the pivot point, and external disturbances. It is important to minimize these factors to obtain accurate results.

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