A mass M = 0.435 kg moves with an initial speed v = 3.43 m/s on a level friction

Question

A mass M = 0.435 kg moves with an initial speed v = 3.43 m/s on a level frictionless air track. The mass is initially a distance D = 0.245 m away from a spring with k = 811 N/m, which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance d, before reversing direction. After bouncing off the spring, the mass travels with the same speed v, but in the opposite direction.

The maximum distance the spring is compressed is: 0.08m

A) Find the total elapsed time until the mass returns to its starting point.

***HINT***
-The mass undergoes a partial cycle of simple harmonic motion while in contact with the spring.
-Total elapsed time means the time from x = D to maximum compression and back to x = D.

Explanation / Answer

The harmonic motion 's time constant. T=2pi/w where w=sqrt(k/m) so T=0.146(s). the partial cycle of simple harmonic motion of the car is half of the full circle. so t=T/2 t=0.073(s). time the the car spent of the linear path. t'=2*0.245/3.43=0.143(s) total time needed. t'+t=0.216(s)

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