1.

What is the maximum acceleration of the particle doing the SHM `gamma=2sin[(pit)/2phi]` where gamma is in cm?A. `(pi)/(2)cm//s^(2)`B. `(pi^(2))/(2)cm//s^(2)`C. `(pi)/(4)cm//s^(2)`D. `(pi)/(4)cm//s^(2)`

Answer» Correct Answer - B
`y=2 sin ((pit)/(2)+phi)`
Comparing the equation with the standard equation
`y=A sin (omegat+phi)`
So `A=2 cm, omega=(pi)/(2)`
Acceleration of particle is
`a=omega^(2)x" "["numerically"]`
At `x=+A, a=a_("max")
`therefore" "a_("max")=omega^(2)A=((pi)/(2))^(2)xx2`
`=2xx(pi^(2))/(4)=(pi^(2))/(2)cm//s^(2)`


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