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 82kg kangaroo hits the ground, the tendons stretch to a maximum length, and the rebound causes the kangaroo to leave the ground approximately 0.15s after its feet first touch.

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


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

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?

Explanation / Answer

1) the period of the motion is twice as big as 0.15 sec. it is 0.3 sec! T=0.3 sec 2) for the simple harmonic oscillator ?=sqrt(k/m) --> k=?2m=(2pf)2m=4p2m/T2=35969 3) the amplitude will increase the time will remain the same

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