A) 0.25 s
B) 4.0 s
C) 1.57 s
D) 3.14 s
Correct Answer: B
Solution :
Maximum acceleration \[=A{{\omega }^{2}}\] \[\Rightarrow \] \[{{\left( \frac{\pi }{2} \right)}^{2}}=1{{\left( \frac{2\pi }{T} \right)}^{2}}\] \[\left( \because 1.57=\frac{\pi }{2} \right)\] \[\Rightarrow \] \[\frac{{{\pi }^{2}}}{4}=\frac{4{{\pi }^{2}}}{{{T}^{2}}}\] \[\Rightarrow \] \[{{T}^{2}}=\frac{4\times 4{{\pi }^{2}}}{{{\pi }^{2}}}\] \[\Rightarrow \] \[{{T}^{2}}=16\] \[\Rightarrow \] T = 4 s
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