Question

A body of mass2kg is suspended from spring of negligible mass and is found to stretch the spring 0.1m.calculate force constant and time period

Answer 1

Answer:

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Answer 2

The force constant and time period are 400 N/m and $\bold{\frac{\pi}{10} \sqrt{2}}$

Force constant:

Let the spring constant be 'k'

When the spring is suspended due to weight and it gets lower, then the gravitational force acting on the spring decreases and potential energy of the spring increases.

Potential energy:

When the spring is compressed or stretched by 'Δx' is:

$U=\frac{1}{2} k(\Delta x)^{2}$

Momentum:

When the spring is stretched maximum, then gravitational potential energy decreases.

$U_{g}=m g \Delta x$

According to law of conservation of energy,

$m g \Delta x=\frac{1}{2} k(\Delta x)^{2}$

$k=\frac{2 m g}{\Delta x}$

On substituting the known values, we get,

$k=\frac{2 \times 2 \times 10}{0.1}$

$\therefore k=400 \ \mathrm{Nm}^{-1}$

Time period:

Time period of a spring mass system is given by the formula,

$T=2 \pi \sqrt{\frac{m}{k}}$

$T=2 \pi \sqrt{\frac{2}{400}}$

$T=2 \pi \sqrt{\frac{2}{(20)^{2}}}$

$T=\frac{2 \pi \sqrt{2}}{20}$

$\therefore T= \frac{\pi}{10} \sqrt{2}$