Understanding the Concepts
* SHM: In SHM, the displacement (x) of an object from its equilibrium position is sinusoidal with time.
* Vmax: The maximum velocity of the object in SHM.
* Relationship between Velocity and Displacement: The velocity (v) in SHM is related to displacement (x) by the equation:
* v = ±ω√(A² - x²)
* where:
* ω is the angular frequency of the oscillation
* A is the amplitude of the oscillation
Finding the Position (x) where v = Vmax/2
1. Start with the velocity equation: v = ±ω√(A² - x²)
2. Set v = Vmax/2: Vmax/2 = ±ω√(A² - x²)
3. Solve for x:
* Square both sides: (Vmax/2)² = ω²(A² - x²)
* Rearrange: x² = A² - (Vmax/2)² / ω²
* Take the square root of both sides (we want the positive position): x = √(A² - (Vmax/2)² / ω²)
Important Notes:
* Angular Frequency (ω): ω = 2πf, where f is the frequency of the oscillation.
* Vmax: Vmax = ωA (maximum velocity in SHM)
* Quadrants: The solution you find represents the positive position. There will also be a corresponding negative position in the opposite direction from the equilibrium point.
Example
Let's say you have an SHM with:
* Amplitude (A) = 5 cm
* Frequency (f) = 2 Hz
To find the positive position where the velocity is half the maximum velocity:
1. Calculate ω: ω = 2πf = 2π(2 Hz) ≈ 12.57 rad/s
2. Calculate Vmax: Vmax = ωA ≈ 12.57 rad/s * 5 cm ≈ 62.85 cm/s
3. Substitute into the equation:
x = √(A² - (Vmax/2)² / ω²)
x ≈ √(5² - (62.85/2)² / 12.57²) ≈ 4.33 cm
Therefore, the positive position where the velocity is half the maximum velocity is approximately 4.33 cm from the equilibrium point.
