A) 1.59mm
B) 1.00m
C) 10 m
D) none of the above
Correct Answer: A
Solution :
: \[T=1.00\times {{10}^{-5}}\sec \] \[{{v}_{m}}=1.00\times {{10}^{3}}m/s\] In S.H.M, maximum speed \[({{v}_{m}})=a\omega =\frac{a\times 2\pi }{T}\] \[\therefore \] \[T={{v}_{m}}=a\times 2\pi \] Or \[a=\frac{T{{v}_{m}}}{2\pi }=\frac{({{10}^{-5}})({{10}^{3}})}{2\times 3.14}=\frac{{{10}^{-2}}}{6.28}m\] Or \[a=\frac{10}{6.28}mm\] Or \[a=1.59\,mm\] Maximum displacement \[(a)=1.59\text{ }mm\]
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