Advertisements
Advertisements
Question
A particle executing S.H.M. has amplitude 0.01 m and frequency 60 Hz. The maximum acceleration of the particle is ____________.
Options
`60 pi^2 "m"//"s"^2`
`88 pi^2 "m"//"s"^2`
`140 pi^2 "m"//"s"^2`
`144 pi^2 "m"//"s"^2`
MCQ
Fill in the Blanks
Solution
A particle executing S.H.M. has amplitude 0.01 m and frequency 60 Hz. The maximum acceleration of the particle is `144 pi^2 "m"//"s"^2`.
Explanation:
The maximum acceleration of a particle executing S.H.M. is given by,
`"a"_"max" = omega^2 "A"`
`"Since," omega = 2 pi"v","therefore we have"`
`"a"_"max" = (2 pi "v")^2 "A"`
Substituting v = 60 Hz; A = 0.01, we get
`"a"_"max" = 144 pi^2 "ms"^-2`
shaalaa.com
Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.
Is there an error in this question or solution?
