The velocity of a particle executing SHM varies with displacement (x) as 4 v²=50-x². The time period of oscillations is x/7 s. The value of x is ___________. ( Take π=22/7)

Solution

Physics sir · 3 steps

Get v² alone first, then match it with v² = ω²(A² - x²) and read off ω.

1. Standard form. In SHM, speed and displacement are linked by v² = ω²(A² - x²). The coefficient in front of the bracket is ω².
2. Isolate v squared. Divide the given relation by 4: v² = 1/4(50 - x²). Comparing, ω² = 1/4, so ω = 1/2 rad s⁻¹ (and A² = 50).

   T = 2π/ω = 2π/(1/2) = 4π = 4 × 22/7 = 88/7 s

3. So. The period is 88/7 s. The question writes it as x/7 s, where this x is just the unknown number, not the displacement. So x = 88.

The common slip: 22. Reading the 4 in front of v² as ω² gives ω = 2, T = π = 22/7 s and so 22. The 4 must be moved across first, which makes ω² = 1/4.

The idea to keep. Whenever v and x are related, put it in the form v² = ω²(A² - x²) with v² standing alone. The number multiplying the bracket is ω², not ω.

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