When a body of mass $1.0 \mathrm{~kg}$ is suspended from a certain light spring hanging vertically, its…
When a body of mass $1.0 \mathrm{~kg}$ is suspended from a certain light spring hanging vertically, its length increases by $5 \mathrm{~cm}$. By suspending $2.0 \mathrm{~kg}$ block to the spring and if the block is pulled through $10 \mathrm{~cm}$ and released, the maximum velocity in it in $\mathrm{m} / \mathrm{s}$ is : (Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$ )
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Solution
$m_1=1 \mathrm{~kg}$, extension $l_1=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$
$\therefore \quad m_1 g=k l_1$
$k=$ force constant of the spring
$k=\frac{m_1 g}{l_1}=\frac{1 \times 10}{5 \times 10^{-2}}=200 \mathrm{~N} / \mathrm{m}$
Time period of the block of mass $2 \mathrm{~kg}$,
$T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{2}{200}}$
$=2 \pi \times \frac{1}{10}=\frac{\pi}{5} \mathrm{~s}$
Maximum velocity, $v_{\max }=A \omega$
where, $\quad A=$ amplitude
$=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}$
$v_{\max }=A \times \frac{2 \pi}{T}=10 \times 10^{-2} \times \frac{2 \pi}{\pi / 5}$
$=10^{-1} \times 2 \times 5$
$=1 \mathrm{~m} / \mathrm{s}$