An assembly of identical spring mass systems is placed on a smooth horizontal surface as shown. Initially springs are non-deformed and masses just touch each other. The left mass is displaced to the left and right mass is displaced to the right by equal and small distances and released. The collisions are elastic. The time period of the oscillation of the system is:
Answers
Answered by
8
In this spring is compressed not expand. Amplitude is half then time period is half
Answered by
17
The time period of the oscillation of the system is π√m/k.
If there is no collision, then each spring will oscillate with period -
T = 2 π √m/k
Due, to the collisions the springs are compressed, however but they can't extend beyond the natural length. Therefore, only half oscillation will be performed.
Thus,
T = 2π√m/k + 2
T = π√m/k
Therefore, the time period of the oscillation of the system is π√m/k.
Similar questions