A) \[60{{\pi }^{2}}m/{{s}^{2}}\]
B) \[88\,{{\pi }^{2}}m/{{s}^{2}}\]
C) \[140\,{{\pi }^{2}}m/{{s}^{2}}\]
D) \[144\,{{\pi }^{2}}m/{{s}^{2}}\]
Correct Answer: D
Solution :
Here: Amplitude a = 0.01 m Frequency\[f\]= 60 Hz Using the relation for maximum acceleration \[=a{{\omega }^{2}}=a{{(2\pi f)}^{2}}=a(4{{\pi }^{2}}{{f}^{2}})\] \[=0.01\times {{\pi }^{2}}\times 4\times {{(60)}^{2}}=144\,{{\pi }^{2}}\,m/{{s}^{2}}\]
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