1) Suppose in an experiment you calculated the period of an oscillation to be 2)
ID: 2279190 • Letter: 1
Question
1) Suppose in an experiment you calculated the period of an oscillation to be
2) Plotting the graph of Period Squared (T^2) vs (Total mass) of an vertical mass-spring oscillation allows you to calculate the spring constant k from the slope of the graph.
3) You and some friends go bungee jumping, but you are afraid of oscillations which go too fast. Which one of these options would provide the slower oscillations?
4) You and some friends go bungee jumping. If two of you decided to share the same bungee cord, would your oscillations be faster or slower than if just one of you used that same cord?
Please don't just guess on these questions.
I really need help figuring them out, so a little blurb on how you figured them out would be really helpful! Thanks.
T = 1.234 plusminus 0.006s You need to calculate the square-root of the period. Your answer is: Suppose instead you plot a graph of T versus . What is the slope of your graph?Explanation / Answer
1) T = T0 + dT =1.234 +/-0.006
dT is the differential of T
sqrt(T) = sqrt(T0) + d[sqrt(T)] = 1.11085 +/- d [sqrt(T)]
if we take the square root from T its differential will be
d [sqrt(T)] =dT/(2*sqrt(T)) = 0.006/(2*1.11085) = 0.0027
The correct answer is thus D)
sqrt(T) =1.111 +/- 0.003 sqrt(s)
2) T = 2*pi*sqrt(m/k) = [2*pi / sqrt(k)]* sqrt(m) = slope* sqrt(m)
if we make a graph of T function of sqrt(m) the slope will be
slope = 2*pi/ [sqrt(k)]
Correct anser is A)
3) T= 2*pi*sqrt(m/k)
for two parallel bungee cords the equlvalent stiftness is K = K1+K2 so its greater than for a single cord.
see http://en.wikipedia.org/wiki/Series_and_parallel_springs#Equivalent_spring
Greater K means smaller T, means faster oscillations.
The correct answer is A) use a single cord.
4) T = 2*pi*sqrt(m/k)
Two persons means increasing mass m, means increasing T, means slower oscillations.
Correct answer is B)
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