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 ________.

Step 2: Calculate the initial velocity at x = 5 cm:

v = ω √(10² - 5²) = ω √(100 - 25) = ω √75 = ω × 5√3

Step 3: Determine the new velocity after it is tripled:

v' = 3v = 3 × (ω × 5√3) = 15ω√3

Step 4: Express the new velocity in terms of the new amplitude A':

v' = ω √(A'² - x²)

Substituting the known values:

15ω√3 = ω √(A'² - 5²)

Step 5: Solve for the new amplitude A':

Square both sides to eliminate the square root:

(15ω√3)² = (ω √(A'² - 25))²

Simplify:

675ω² = ω² (A'² - 25)

Divide both sides by ω²:

675 = A'² - 25

Add 25 to both sides:

A'² = 700

Take the square root of both sides:

A' = √700

Simplify √700:

√700 = √(7 × 100) = 10√7