Question

please work 21 and 22. The slope of wooden vall graph is 3.916. Slope of metal ball graph is 4.052.

Lis e d from the 15. are the mass of the metal ball 416 pts) 16. Construct a simple pendam 100.0 cm in length in the metal alla som si center of man of the ball) 17. Move the pendam from equilibrium (about 10.20) and release Merethe time 18. Determine the period. Record it in the table provided T=.991 19. Shorten L in increments of 200 cm and measure T for each length 20. Repeat the procedure using the wooden ball. Mas of the wooden ball. m .O0 for 20 cm 1 2 pts) Period of Oscillation: 982 0 cm Metal 2.015 Wood 1.995 1.80s 154s 1zes 1.1.535 1,23 1,235 .904s .9175 o o sl 9 Graph) Error Metal 21. Produce graphs of 7° . I for each ball. Apply a linear fit; print a copy for each partner (6 pts). Squaring both sides of the simple pendulum equation, it is apparent that the slope of at us. I graph should be equal to 47/9. From the slope of the graphs, calculate 9. Show work. (10 pts) Wood T= ang T = 1 5 - (2) - ģ-g.(1)² = 2 A) 22. Consider the pendulums pictured at the right. Describe how one would de termine how much more time pendulum A takes to travel a distance 2r than for pendulum B to travel 1, given they have the same length. Assume the pendulums are at rest at the positions in the figures, then released. (5 pts)

Answer 1

21. The time period, T is given by

$T=2\pi\sqrt{\frac{L}{g}},\ or\ T^{2}=\frac{4\pi^{2}L}{g}=\frac{4\pi^{2}}{g}L$

If you plot T² as y and L as x then the slope (y = mx + c) will be

$Slope=\frac{4\pi^{2}}{g},\ or\ g=\frac{4\pi^{2}}{slope}=\frac{39.478}{slope}$

Therefore, the slope for metal, g = 39.478/4.052 = 9.743 m/s² and for wood, g = 39.478/3.916 = 10.081 m/s².

Let us consider the value of g is 9.81 m/s² and then the percentage error will be

$\%Error=\frac{\Delta g}{g}\times 100$

Material	g (Standard)	g (Graph	% Error
Metal	9.81	9.743	-3.44%
Wood	9.81	10.081	+2.76%

22. The time period, T is given by

$T=2\pi\sqrt{\frac{L}{g}}$

The time period is not dependent on the displacement. Hence for ideal pendulum, you will obtain same period for two case.

But in real pendulum, the pendulum the amplitude will decay with time and after some oscillation the pendulum will come to rest. For more amplitude at the starting position, the decay process will faster. It may be come to rest before 20 oscillation. Then you can not measure the time period of the pendulum accurately as you were asked to measure time for 20 oscillation to determine the period (T).

On the other hand if you have starting amplitude less, then it will behave close to ideal pendulum. It will oscillate more before it will come to rest. You can measure easily 20 equal oscillation without decaying the amplitude much.

Therefore, if you start the pendulum as shown in B, you will obtain more accurate time period hence the value of g.