A particle performs harmonic oscillations along a straight line with a period of 6s and amplitude 4 cm. The mean velocity of the particle averaged over the time interval during which it travels a distance of 2 cm starting from the extreme position is

Options

A. `1cms^(-1)`
B. `2cms^(-1)`
C. `4cms^(-1)`
D. `8cms^(-1)`

Answer

B

Explanation

Here, `T=6s, r=4cm,` Displacement from the mean position when the particle travels 2 cm from the extreme position is,`x=4-2=2.0cm. ` If t is the time taken then `x=r cos omega t =r cos ((2pi)/(T))t`
`:. 2=4cos ((2pi)/(6))t or cos ((pit)/(3))=(2)/(4)=(1)/(2)=cos((pi)/(3))`
so
`:. ` Average velocity `=` `(dist ance)/(time )=(2)/(1)=2cms^(-1)`