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 √x cm. The value of x is _____________.

Solution

Physics sir · 3 steps

The jet changes only the speed, not the position or ω, so rebuild the amplitude from the new speed.

1. Speed at 5 cm. In SHM, v = ω√(A² - x²). With A = 10 cm and x = 5 cm, v = ω√(100 - 25) = ω√75.
2. What stays fixed. The air jet acts at that instant, so the body is still at x = 5 cm. The angular frequency ω = √(k/m) depends only on the system, so it does not change. Only the speed becomes 3ω√75.

   A'² = x² + v'²/ω² = 25 + 9 × 75 = 700

3. So. A' = √700 cm, so x = 700.

The common slip: 675. Leaving out the x² = 25 term and using A'² = v'²/ω² alone gives 675, not 700. The body is 5 cm from the mean position, so that term must stay.

The idea to keep. A sudden kick in SHM keeps x and ω the same. Use A² = x² + v²/ω² with the new speed. Tripling the speed triples only the v²/ω² part, so square the 3.

Answer from the NTA final answer key. Solution by PrepWiser.