When a body of mass $1.0\, kg$ is suspended from a certain light spring hanging vertically, its length increases by $5\, cm$. By suspending $2.0\, kg$ block to the spring and if the block is pulled through $10\, cm$ and released the maximum velocity in it in $m/s$ is : (Acceleration due to gravity $ = 10\,m/{s^2})$
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(b) 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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