In simple harmonic motion (SHM), the relationship between displacement (x) and velocity (v) is given by:
v = ω√(A² - x²)
where ω is the angular frequency and A is the amplitude of the motion.
To find the displacement at which the velocity is half of its maximum value, we need to determine the maximum velocity (v_max) and then solve for x when v = v_max/2.
The maximum velocity occurs at the amplitude of the motion, so v_max = ω√(A² - 0²) = ωA.
Now, let's solve for x when v = v_max/2:
v = v_max/2 ω√(A² - x²) = (ωA)/2 √(A² - x²) = A/2 A² - x² = (A/2)² A² - x² = A²/4 4A² - 4x² = A² 3A² = 4x² x² = (3A²)/4 x = ±√((3A²)/4)
Therefore, the displacement from the mean position where the velocity is half of its maximum value is x = ±√((3A²)/4). The positive and negative signs indicate the two possible directions of displacement.