Question

When a body of mass $1.0\text{kg}$ is suspended from a certain light spring hanging vertically, its length increases by $5\text{cm}$. If by replacing the body of mass $1.0\text{kg}$ and suspending $2.0\text{kg}$ block to the spring, the block is pulled through $10\text{cm}$ from its new equilibrium position and released, the maximum velocity achieved by the block is

• A. $0.5\text{m/s}$
• B. $1\text{m/s}$
• C. $2\text{m/s}$
• D. $4\text{m/s}$

Answer 1

The correct option is B $1\text{m/s}$

When the body of mass $1\text{kg}$ is suspended, extension in spring i.e $x=5\text{cm}=0.05\text{m}$
Applying equilibrium condition,
Force applied by spring = Gravitational force
$kx=mg$
$\Rightarrow k\left(0.05\right)=10\Rightarrow k=200\text{N/m}$
Now a body of mass $2\text{kg}$ is suspended from same spring and pulled by $10\text{cm}$ from its equillibrium position.
$\Rightarrow$ Amplitude of oscillation $\left(A\right)=10\text{cm}=0.1\text{m}$
The block will perform SHM,
$\Rightarrow V_{max}=A\omega$
where $\omega =\sqrt{\frac{k}{m}}$
$\therefore V_{max}=0.1\times \sqrt{\frac{200}{2}}=1\text{m/s}$