The jumping gait of the kangaroo is efficient because energy is stored in the st

Question

The jumping gait of the kangaroo is efficient because energy is stored in the stretch of stout tendons in the legs; the kangaroo literally bounces with each stride. We can model the bouncing of a kangaroo as the bouncing of a mass on a spring. A 84 kangaroo hits the ground, the tendons stretch to a maximum length, and the rebound causes the kangaroo to leave the ground approximately 0.14 after its feet first touch.

A) Modeling this as the motion of a mass on a spring, what is the period of the motion?

B) Given the kangaroo mass and the period you've calculated, what is the spring constant?

C) If the kangaroo speeds up, it must bounce higher and farther with each stride, and so must store more energy in each bounce. How does this affect the time and the amplitude of each bounce?
1 amplitude increase, time remains the same
2 amplitude decrease, time decrease
3 amplitude increase, time increase
4 amplitude increase, time decrease
5 amplitude remains the same, time increase
6 amplitude decrease, time increase

Explanation / Answer

The period T of the whole motion must be twice the value for the half at the bottom. So T is 0.2 sec w ( the angular frequency) is 2Pi()/T Now k ( the spring constant) = F/R = m w^2 = 84* (2*pi()/0.14)^2 = 16.9 * 10 ^4 N/m finally f. The period for SHM is not affected by the amplitude of the bounce. 5 amplitude remains the same, time increase

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