A) \[60{{\pi }^{2}}\,m/{{s}^{2}}\]
B) \[80{{\pi }^{2}}\,m/{{s}^{2}}\]
C) \[120{{\pi }^{2}}\,m/{{s}^{2}}\]
D) \[144{{\pi }^{2}}\,m/{{s}^{2}}\]
Correct Answer: D
Solution :
In S.H.M., maximum acceleration\[{{a}_{\max }}\] and amplitude A are related as \[{{a}_{\max }}={{\omega }^{2}}A={{(2\pi n)}^{2}}A\] \[=4{{\pi }^{2}}{{n}^{2}}A\] Here: n = 60 Hz, A = 0.01m \[\therefore \] \[{{a}_{\max }}=4{{\pi }^{2}}\times {{(60)}^{2}}\times (0.01)\] \[=144\pi m/{{s}^{2}}\]
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