1. Initial SHM data

• Initial amplitude: $A = 10\text{ cm}$
• At the instant considered, displacement from mean position: $x = 5\text{ cm}$

For a particle in SHM, speed at displacement $x$ is

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

So initially,

$v_1 = \omega\sqrt{10^2 - 5^2} = \omega\sqrt{100 - 25} = \omega\sqrt{75} = 5\omega\sqrt{3}$

2. Velocity is tripled

The air jet triples the speed instantly, so the new speed becomes

$v_2 = 3v_1 = 15\omega\sqrt{3}$

At that instant, the displacement is still $x=5\text{ cm}$.

3. Find the new amplitude

After the velocity change, the particle continues in SHM with the same angular frequency $\omega$ and new amplitude $A'$.

Using the SHM relation again:

$v_2^2 = \omega^2\left(A'^2 - x^2\right)$

Substitute values:

$(15\omega\sqrt{3})^2 = \omega^2(A'^2 - 25)$

$675\omega^2 = \omega^2(A'^2 - 25)$

$A'^2 - 25 = 675$

$A'^2 = 700$

Thus,

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

Since the new amplitude is given as $\sqrt{x}\text{ cm}$,

$x = 700$

4. Comparison with stored answer

Derived answer: $700$

Stored correct answer: $700$

They match.