A particle executing simple harmonic motion is given by  $y=8\sin \pi \left(10t-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$ where y  is in metre, and t is in seconds. Then the average speed of the particle during one  full oscillation is 

detailed solution

Correct option is D

Average speed of a particle executing simple harmonic motion = Total   DistanceTotal   time   In a simple harmonic motion, the total distance travelled by the particle is 4 times  the amplitude. The total time taken is one full time period of oscillation.Therefore,  Average    speed=Total   DistanceTotal   time=4ATComparing the given equation y=8sinπ10t−12 with standard equation,  y=Asinωt+ϕ , we get amplitude=  A=8  m and angular velocity= ω=10π .Using the relation ω=2πT ,substitute  ω=10π we get 10π=2πT  i.e. T=15  s.=time periodHence  Average    speed=4AT=4×815=160  m/s.