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A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is[2007]
- a)T/8
- b)T/12
- c)T/2
- d)T/4
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A particle executes simple harmonic oscillation with an amplitude a. T...
Displacement from the mean position
According to problem y = a/2
This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.
Most Upvoted Answer
A particle executes simple harmonic oscillation with an amplitude a. T...
Understanding Simple Harmonic Motion (SHM)
In simple harmonic motion, a particle oscillates around an equilibrium position with a specific amplitude and period. Key parameters include:
- Amplitude (a): The maximum displacement from the equilibrium position.
- Period (T): The time taken for one complete cycle of motion.
Objective of the Problem
We need to find the minimum time taken by the particle to travel half of the amplitude (a/2) from the equilibrium position.
Position Equation in SHM
The position of the particle in SHM can be described by the equation:
- x(t) = a * sin(ωt)
where ω = 2π/T (angular frequency).
Finding the Time to Reach Half Amplitude
To find the time taken to reach a displacement of a/2:
- Set x(t) = a/2:
a/2 = a * sin(ωt)
- Simplifying gives:
1/2 = sin(ωt)
- This implies:
ωt = π/6 (since sin(π/6) = 1/2)
Calculating Time
Now, we can find t:
- t = (π/6) / ω
- Substitute ω = 2π/T:
t = (π/6) / (2π/T) = T/12
Conclusion
The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is T/12. Thus, the correct answer is option B: T/12.
In simple harmonic motion, a particle oscillates around an equilibrium position with a specific amplitude and period. Key parameters include:
- Amplitude (a): The maximum displacement from the equilibrium position.
- Period (T): The time taken for one complete cycle of motion.
Objective of the Problem
We need to find the minimum time taken by the particle to travel half of the amplitude (a/2) from the equilibrium position.
Position Equation in SHM
The position of the particle in SHM can be described by the equation:
- x(t) = a * sin(ωt)
where ω = 2π/T (angular frequency).
Finding the Time to Reach Half Amplitude
To find the time taken to reach a displacement of a/2:
- Set x(t) = a/2:
a/2 = a * sin(ωt)
- Simplifying gives:
1/2 = sin(ωt)
- This implies:
ωt = π/6 (since sin(π/6) = 1/2)
Calculating Time
Now, we can find t:
- t = (π/6) / ω
- Substitute ω = 2π/T:
t = (π/6) / (2π/T) = T/12
Conclusion
The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is T/12. Thus, the correct answer is option B: T/12.
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A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is[2007]a)T/8b)T/12c)T/2d)T/4Correct answer is option 'B'. Can you explain this answer?
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