A particle of mass $$0.50 \text{ kg}$$ executes simple harmonic motion under force $$F = -50 \text{ (Nm}^{-1}\text{)}x$$. The time period of oscillation is $$\frac{\pi}{x}$$ s. The value of $$x$$ is ______ (Given $$\pi = \frac{22}{7}$$)

A particle of mass $$m = 0.50$$ kg executes SHM under force $$F = -50x$$ N. We need to find the value of $$x$$ if the time period is $$\frac{\pi}{x}$$ s.

Comparing with $$F = -kx$$, we get $$k = 50$$ N/m.

$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{0.50}} = \sqrt{100} = 10 \text{ rad/s}$$

$$T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{\pi}{5} \text{ s}$$

Given $$T = \frac{\pi}{x}$$, comparing with $$T = \frac{\pi}{5}$$:

$$x = 5$$

The answer is 5.