Consider a spring that exerts restoring force $F = - k x$ for $x > 0, F = - 2 k x$ for $x < 0$ . A mass $m$ on a frictionless surface is attached to this spring, displaced to $x = A$ and released. Find the time period of motion.

5.36 \sqrt{\frac{m}{k}}

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5.36 \sqrt{\frac{m}{2 k}}

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5.36 \sqrt{\frac{2 m}{k}}

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Solution

The correct option is A $5.36 \sqrt{\frac{m}{k}}$
Given,
$F = - k x, x > 0$
and $F = - 2 k x, x < 0$

For half time-period,
$F_{net} = - k x$ $∴ a = (\frac{- k}{m}) x$
As $T = \frac{2 π}{ω}$ , hence $t_{1} = \frac{2 π}{2 \sqrt{k / m}} = \frac{π}{\sqrt{k / m}}$

Again, for other half time-period,
$F_{net} = - 2 k x$ $∴ a = (\frac{- 2 k}{m}) x$
As $T = \frac{2 π}{ω}$ , hence $t_{2} = \frac{2 π}{2 \sqrt{2 k / m}} = \frac{π}{\sqrt{2 k / m}}$

Hence, the total time period is
$T = t_{1} + t_{2} = \frac{π}{\sqrt{k / m}} + \frac{π}{\sqrt{2 k / m}} = 5.36 \sqrt{m / k}$

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