A particle executes simple harmonicmotion with an amplitude of 5 cm.When the particle is at 4 cm from themean position, the magnitude of itsvelocity in SI units is equal to that of itsacceleration. Then, its periodic time inseconds is
Question
A particle executes simple harmonicmotion with an amplitude of 5 cm.When the particle is at 4 cm from themean position, the magnitude of itsvelocity in SI units is equal to that of itsacceleration. Then, its periodic time inseconds is
Solution
In simple harmonic motion, the velocity and acceleration of a particle are given by:
v = ω√(A² - x²) and a = ω²x
where:
- v is the velocity
- ω is the angular frequency
- A is the amplitude
- x is the displacement
- a is the acceleration
Given that the velocity equals the acceleration, we can set the two equations equal to each other and solve for ω:
ω√(A² - x²) = ω²x √(A² - x²) = ωx ω = √(A² - x²) / x
Substituting the given values A = 5 cm = 0.05 m and x = 4 cm = 0.04 m:
ω = √((0.05)² - (0.04)²) / 0.04 ω = √(0.0025 - 0.0016) / 0.04 ω = √0.0009 / 0.04 ω = 0.03 / 0.04 ω = 0.75 s⁻¹
The period T of the motion is given by T = 2π/ω. Substituting the calculated value of ω:
T = 2π / 0.75 T = 8.37758 seconds
So, the periodic time of the particle's motion is approximately 8.38 seconds.
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