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A particles in simple harmonic motion has a time period of 2 second if the amplitude of motion is 0.5 m, find the velocity of the particles when its displacement is 0.3m?
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A particles in simple harmonic motion has a time period of 2 second if...
Understanding Simple Harmonic Motion
Simple Harmonic Motion (SHM) describes the oscillatory motion of a particle about an equilibrium position. The key parameters in SHM are amplitude, time period, and displacement.
Given Data
- Time Period (T): 2 seconds
- Amplitude (A): 0.5 m
- Displacement (x): 0.3 m
Formulas to Use
1. The velocity (v) in SHM can be calculated using the formula:
v = ω √(A² - x²)
2. The angular frequency (ω) is given by:
ω = 2π/T
Calculating Angular Frequency
- Substitute the time period into the formula:
ω = 2π/2 = π rad/s
Calculating Velocity
- Next, substitute the values into the velocity formula:
- A² = (0.5)² = 0.25 m²
- x² = (0.3)² = 0.09 m²
- Therefore, A² - x² = 0.25 - 0.09 = 0.16 m²
- Now, calculate the velocity:
v = π √(0.16) = π * 0.4 = 1.256 m/s (approximately)
Conclusion
The velocity of the particle when its displacement is 0.3 m is approximately 1.26 m/s. This value indicates the speed at which the particle moves through its oscillation at that specific point in its cycle. Understanding these calculations helps in analyzing the dynamics of SHM effectively.
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A particles in simple harmonic motion has a time period of 2 second if the amplitude of motion is 0.5 m, find the velocity of the particles when its displacement is 0.3m?
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