When a body of mass 1.0, kg is suspended from a certain light spring hanging vertically, its length

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13.Oscillations

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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})$

A

$0.5$

B

$1$

C

$2$

D

$4$

Solution

(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, ${v_{\max }} = a\omega = (10 \times {10^{ – 2}}) \times 10 = 1\,m/s$

Standard 11

Physics

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