Question

When a body of mass $1.0kg$ is suspended from a certain light spring hanging vertically, its length increases by $5cm$. By suspending $2.0kg$ block to the spring and if the block is pulled through $10cm$ and released, the maximum velocity in it in $m/s$ is

• A. $0.5$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (B)

$1$

Using equilibrium $kx=10$
$\Rightarrow k=10\times 100$ $=200N/m$
Extension $=10cm=x_1$
Velocity will be max when block passes through equilibrium position i.e P.E of spring changes to K.E
$\pm$ using S.H.M
where, $h=10+10\left(\sin \omega t\right)$

Maximum speed $=10\omega$

$=10\times \sqrt{\frac{k}{m}}$

$=10\times \sqrt{\frac{k}{m}}\times {10}^{-2}$

$=1m/s$