A particle of mass $$250$$ g executes a simple harmonic motion under a periodic force $$F = (-25x)$$ N. The particle attains a maximum speed of $$4$$ m s$$^{-1}$$ during its oscillation. The amplitude of the motion is ______ cm.

Solution :

Given restoring force :

$$F = -25x$$

Comparing with :

$$F = -kx$$

we get,

$$k = 25\text{ N m}^{-1}$$

Mass of particle :

$$m = 250\text{ g}$$

$$= 0.25\text{ kg}$$

Angular frequency :

$$\omega = \sqrt{\frac{k}{m}}$$

$$= \sqrt{\frac{25}{0.25}}$$

$$= \sqrt{100}$$

$$= 10\text{ rad s}^{-1}$$

Maximum speed in SHM is :

$$v_{max} = A\omega$$

Given,

$$v_{max} = 4\text{ m s}^{-1}$$

Therefore,

$$A = \frac{v_{max}}{\omega}$$

$$= \frac{4}{10}$$

$$= 0.4\text{ m}$$

$$= 40\text{ cm}$$

Final Answer :

$$40\text{ cm}$$