A) 0.5
B) 1
C) 2
D) 4
Correct Answer: B
Solution :
Initially when 1 kg mass is suspended then by using \[F=kx\] \[\Rightarrow \] \[mg=kx\]\[\Rightarrow \]\[k=\frac{mg}{x}=\frac{1\times 10}{5\times {{10}^{-2}}}=200\frac{N}{m}\] Further, the angular frequency of oscillation of 2 kg mass is \[\omega =\sqrt{\frac{k}{M}}=\sqrt{\frac{200}{2}}=10\,rad/sec\] Hence, \[{{v}_{\max }}=a\omega =(10\times {{10}^{-2}})\times 10=1\,m/s\]
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