Is there a significant difference between the two slopes in the acceleration vs sin(theta) graph? What quantity does the slope represent in this experiment?
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Is there a significant difference between the two slopes in the acceleration vs sin(theta) graph? What quantity does the slope represent in this experiment?
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- You may also recall that for a simple pendulum N = V, where g is the acceleration due to gravity and L is the length of the pendulum. This has been inserted in Equation 4 above. Lastly, this means that angular position 0 of the pendulum is given by: 0 = 0, cos(Nt) Equation 5 Question 1 Use the information above and Equation 2 to write an equation for the period of the pendulum in terms of g and L.A pendulum has a period of 5.8 s. If you were to transfer this pendulum to the surface of Mars (where g is 3.71 meters per second squared), what would the period (in seconds) be?I am having a hard time finding the meaning of what T2 is in the equation. Do I just square the T or do something else? Thanks for the help in advanced.
- The quantities A and φ (called the amplitude and the phase) are undetermined by the differential equation. They are determined by initial conditions -- specifically, the initial position and the initial velocity -- usually at t = 0, but sometimes at another time. In the oscillating part of the experiment, I measured only the time of 30 periods. I measured no position or velocity. Consequently, A and φ (and also y0) are irrelevant in the problem. We only compare the period T or the frequency ω with the theoretical prediction. You have (hopefully) derived (or maybe looked up) the relation between ω and k and m. This final question relates ω and T. If ω = 8.2*102 rad/s, calculate T in seconds. (Remember, that a radian equals one.) T might be a fraction of a second.Please help solve for v in (m/s)Note: Because the argument of the trigonometric functions in this problem will be unitless, your calculator must be in radian mode if you use it to evaluate any trigonometric functions. You will likely need to switch your calculator back into degree mode after this problem.A massless spring is hanging vertically. With no load on the spring, it has a length of 0.24 m. When a mass of 0.59 kg is hung on it, the equilibrium length is 0.98 m. At t=0, the mass (which is at the equilibrium point) is given a velocity of 4.84 m/s downward. At t=0.32s, what is the acceleration of the mass? (Positive for upward acceleration, negative for downward)