A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is $$\sqrt{x}$$ cm. The value of $$x$$ is ______.

We are given: Amplitude $$A = 10$$ cm. At displacement $$x = 5$$ cm from the mean position, the velocity is tripled by an air jet.

First, the original velocity at $$x = 5$$ cm is determined. In SHM, the velocity at displacement $$x$$ is:

$$v = \omega\sqrt{A^2 - x^2}$$

$$v = \omega\sqrt{10^2 - 5^2} = \omega\sqrt{100 - 25} = \omega\sqrt{75}$$

Next, the new velocity after tripling is:

$$v' = 3v = 3\omega\sqrt{75}$$

Finally, since the angular frequency $$\omega$$ remains the same, the new amplitude $$A'$$ satisfies at displacement $$x = 5$$ cm with velocity $$v'$$:

$$v' = \omega\sqrt{A'^2 - x^2}$$

$$3\omega\sqrt{75} = \omega\sqrt{A'^2 - 25}$$

Dividing both sides by $$\omega$$ gives:

$$3\sqrt{75} = \sqrt{A'^2 - 25}$$

Squaring both sides yields:

$$9 \times 75 = A'^2 - 25$$

$$675 = A'^2 - 25$$

$$A'^2 = 700$$

$$A' = \sqrt{700} \text{ cm}$$

Since the new amplitude is $$\sqrt{x}$$ cm, it follows that $$x = 700$$.

Hence, the value of $$x$$ is 700.