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 :
The maximum acceleration of a particle in SHM is\[a=-{{\omega }^{2}}y\] Where \[\omega \] is angular velocity and \[y\] is displacement. Also, \[\omega =2\pi n\] \[a=-{{\left( 2\pi n \right)}^{2}}y\] Given, \[y=0.01\,m,\,n=60\,Hz\] \[\therefore \] \[a=-{{\left( 2\pi \times 60 \right)}^{2}}\times 0.01\] \[a=144{{\pi }^{2}}\,\,m/{{s}^{2}}\](Leave the negative sign).
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