1.

(i) A particle is performing simple harmonic motion with time period T. At an instant its speed is 60% of its maximum value and is increasing. After an interval `Delta t` its speed becomes 80% of its maximum value and is decreasing. Find the smallest value of `Delta t` in terms of T. (ii) A particle is doing SHM of amplitude 0.5 m and period `pi` seconds. When in a position of instantaneous rest, it is given an impulse which imparts a velocity of `1 m//s` towards the equilibrium position. Find the new amplitude of oscillation and find how much less time will it take to arrive at the next position of instantaneous rest as compared to the case if the impulse had not been applied.

Answer» Correct Answer - (i)` Delta T=(t)/(4), (ii) A=(1)/(sqrt(2))m; t=(pi)/(8) s`


Discussion

No Comment Found

Related InterviewSolutions