Question

A block of mass $m=1\text{kg}$ attached to a massless spring of spring constant $200\text{N/m}$ is passing over a pulley. If we displace the block by $5\text{cm}$ in vertical direction, the block starts oscillation about the equilibrium position. What is the speed of block at its mean point of oscillation ?
Assuming, pulley is massless.

• A. $0.2\text{m/s}$
• B. $0.7\text{m/s}$
• C. $0.5\text{m/s}$
• D. $1.0\text{m/s}$

Answer 1

The correct option is B $0.7\text{m/s}$

Elongation in spring at equilibrium will be

$x_1=\frac{mg}{k}=\frac{10}{200}$

$\therefore x_1=0.05\text{m}$

Now, when we stretch spring about $5\text{cm}=0.05\text{m}$ by displacing a block.

Total mechanical energy $=$ Energy stored in spring $+$ Gravitational potential energy

We assume that, the gravitational potential energy of the block is zero at equilibrium.

Gravitational potential energy = $-mgH$

$U_{GE}=-10\times 0.05=-0.5\text{J}$

Energy stored in spring $=\frac{1}{2}k{x}_{total}^2$

$U_{SE}=\frac{1}{2}K(x+x_1{)}^2$

$=\frac{1}{2}\times 200\times {0.1}^2=1\text{J}$

$\therefore$ Total mechanical energy $=1-0.5=0.5\text{J}$

Now, at mean position (equilibrium position)

From conservation of mechanical eenrgy, total mechanical energy at the mean position will be

$E_{mean}=\frac{1}{2}k{x}_1^2+\frac{1}{2}m{v}^2=0.5\text{J}$

$\frac{1}{2}k{x}_1^2+\frac{1}{2}m{v}^2=0.5$

$\Rightarrow 0.25+\frac{1}{2}{v}^2=0.5$

$\therefore v=\sqrt{0.5}=0.707\text{m/s}$

Hence, option $\left(\text{B}\right)$ is correct.

Why this question ? This question introduce application of energy conservation law in simple harmonic oscillator system. In such problems, find out every potential energy governed the system and kinetic energy of every element at different instants and make them equal.