Unit VIII - Waves/Oscillations/Simple Harmonic Motion

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Exploring Parameter A

A, or amplitude, governs the maximum distance away from the position of equilibrium in any system undergoing simple harmonic motion. We changed the A by releasing the mass hangar from a higher height, in this case 0.10 m. In the equationy=A\sin\left(Bt+C\right)+D, A is shown to be the maximum distance because the sine function varies between -1 and 1 over its period and this output multiplied by A, gives a range of -A+D and A+D. The average of these two points gives the equilibrium position, which happens to be D, and the distance between D and A+D or D and -A+D is A.

Exploring Parameter D

In order to make the D parameter -0.05 m, we moved up the motion detector 5 cm by placing several books between the motion detector and the table. This decreased the distance between the equilibrium position and the motion detector (which used to be calibrated to be 0 m) by 0.05 m, making the equilibrium position now have an absolute y position of -0.05 m. In order to reverse such a graph across time axis. We would have taken the books used to prop up the motion detector 5 cm and placed them under the stand holding the oscillator, thus increasing the distance from the original by 0.05 cm making the new offset 0.05 m.

Exploring Parameter C

Logger Pro was set up to start collecting data when the y-value was increasing across 0 cm, or the equilibrium position, therefore the left to right shift caused by C of the sine function would be very nearly zero due to the fact that the first value of the equilibrium position should occur at nearly 0 s (or at the pt (0,0), the origin of the sine graph). In order to make the graph have a C value of \frac{\pi}{2}, one would have to start the system's data collection at the top of the mass's ascent because this would be 1/4 of the way through the total motion of the mass across one wave cycle (as \frac{\pi}{2}is 1/4 of 2\pi, the period of the sine function). To create a C value of \pi, one would start the data collection when the mass was at the equilibrium position with a descending trajectory, since this would complete 1/2 (since \pi is 1/2 of 2\pithe period of the sine function) of the total wave cycle

Exploring Parameter B

We did not collect what T was but I can say that \frac{T}{2\pi}=B=\omega. When B was changed to 6.28 rad/s, one could rearrange the equation to solve for B: \frac{B}{2\pi}=T.  6.28\:rad\:\approx\:2\pi so T, and thus the period of the sine function, must equal 1 second. In order to double the angular frequency of the system, one would have to make the mass \frac{\sqrt{2}}{2}x that of its original value.

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