L (meters) mass Time for 30 oscillations Period T (seconds) T² g = 39.44L/T2
Average value of g = __________
QUESTIONS:
1) From your data what effect does changing the mass have on the period (for a
given value of the length L)?
2) What role, if any, does air resistance have on your results? Explain your
reasoning.
3) Would you conclude that Galileo was correct in his observation that the
period of a simple pendulum depends only on the length of the pendulum?
4) On the moon, the acceleration due to gravity is one-sixth that of earth. That is
gmoon = gearth /6 = (9.8 m/s2
)/6 = 1.63 m/s2
.
What effect, if any, would this have on the period of a pendulum of length L?
How would the period of this pendulum differ from an equivalent one on earth.
"T^2=39.44\\cdot\\frac{l}{g}"
QUESTIONS:
1) From your data what effect does changing the mass have on the period (for a given value of the length L)?
Answer. The change in mass does not affect the period of oscillation of a simple pendulum.
2) What role, if any, does air resistance have on your results? Explain your reasoning.
Answer. The force of air resistance reduces the amplitude of oscillations.
3) Would you conclude that Galileo was correct in his observation that the period of a simple pendulum depends only on the length of the pendulum?
Answer. In accordance with the exact solution, the period of oscillation of a simple pendulum depends on the length of the pendulum and the angle of deviation. For small angles of deviation, we can assume that the period depends only on the length of the pendulum.
4) On the moon, the acceleration due to gravity is one-sixth that of earth. That is gmoon = gearth /6 = (9.8 m/s2)/6 = 1.63 m/s2 .What effect, if any, would this have on the period of a pendulum of length L? How would the period of this pendulum differ from an equivalent one on earth.
Answer. The period of oscillation of a simple pendulum will increase with increasing acceleration of free fall.
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