A particle executes S.H.M. with amplitude 'a', and time period 'T'. The displacement of the particle when its speed is half of maximum speed is (√x a)/2. The value of x is __________.

Solution

Physics sir · 3 steps

Use v = ω√(a² - y²), set it equal to half of ω a, and solve for y.

1. Speed at displacement y. In SHM the speed at displacement y from the mean position is v = ω√(a² - y²). It is largest at y = 0, where vₘₐₓ = ω a.

   ω√(a² - y²) = (ω a)/2

2. Solve for y. Cancel ω and square both sides: a² - y² = a²/4. So y² = 3a²/4, which gives y = (√3 a)/2.
3. So. Comparing with (√x a)/2 gives x = 3. The time period T is not needed, because ω cancels.

The common slip: 2. Halving v² instead of v gives a² - y² = a²/2, so y = (√2 a)/2 and x = 2. Square the factor 1/2 too, which gives 1/4.

The idea to keep. Whenever speed and displacement are linked in SHM, reach for v = ω√(a² - y²). Half the maximum speed is not at half the amplitude: it is at √3/2a.

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