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A particle performs SHM with a time period T and amplitude 'a'. The magnitude of average velocity of the particle over the time interval during which it travels a distance a/2 from the extreme position is :
- a)a/T
- b)2a/T
- c)3a/T
- d)a/2T
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A particle performs SHM with a time period T and amplitude a. The magn...
The magnitude of displacement in the given time interval = a/2
Time taken by the particle to cover a distance a/2 starting from rest = T/6
Hence the magnitude of average velocity over given time interval is
Time taken by the particle to cover a distance a/2 starting from rest = T/6
Hence the magnitude of average velocity over given time interval is
Most Upvoted Answer
A particle performs SHM with a time period T and amplitude a. The magn...
Explanation:
In Simple Harmonic Motion (SHM), the displacement of a particle from its mean position can be represented by the equation:
x = A sin(ωt + φ)
where,
x = displacement of the particle
A = amplitude of the motion
ω = angular frequency of the motion
t = time
φ = phase constant
The time period (T) of the motion is the time taken for one complete oscillation and is given by:
T = 2π/ω
Calculating Average Velocity:
The average velocity of the particle over a given time interval can be calculated by dividing the total displacement by the total time taken.
In this case, the particle travels a distance of a/2 from the extreme position. Let's find the time taken to travel this distance.
When the particle is at a distance of a/2 from the extreme position, the displacement can be written as:
x = A sin(ωt + φ) = a/2
Solving this equation, we can find the value of ωt + φ.
Now, let's consider the particle at the extreme position. At this point, the displacement is equal to the amplitude A. Using the same equation, we can write:
x = A sin(ωt + φ) = A
Solving this equation, we can find the value of ωt + φ.
The time interval between these two positions is equal to the time period T.
Using the formula for average velocity:
Average Velocity = Total Displacement / Total Time
In this case, the total displacement is a/2 and the total time is T.
Therefore, the magnitude of the average velocity is given by:
Average Velocity = (a/2) / T = a / (2T)
Hence, the correct answer is option C: 3a/T.
In Simple Harmonic Motion (SHM), the displacement of a particle from its mean position can be represented by the equation:
x = A sin(ωt + φ)
where,
x = displacement of the particle
A = amplitude of the motion
ω = angular frequency of the motion
t = time
φ = phase constant
The time period (T) of the motion is the time taken for one complete oscillation and is given by:
T = 2π/ω
Calculating Average Velocity:
The average velocity of the particle over a given time interval can be calculated by dividing the total displacement by the total time taken.
In this case, the particle travels a distance of a/2 from the extreme position. Let's find the time taken to travel this distance.
When the particle is at a distance of a/2 from the extreme position, the displacement can be written as:
x = A sin(ωt + φ) = a/2
Solving this equation, we can find the value of ωt + φ.
Now, let's consider the particle at the extreme position. At this point, the displacement is equal to the amplitude A. Using the same equation, we can write:
x = A sin(ωt + φ) = A
Solving this equation, we can find the value of ωt + φ.
The time interval between these two positions is equal to the time period T.
Using the formula for average velocity:
Average Velocity = Total Displacement / Total Time
In this case, the total displacement is a/2 and the total time is T.
Therefore, the magnitude of the average velocity is given by:
Average Velocity = (a/2) / T = a / (2T)
Hence, the correct answer is option C: 3a/T.
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A particle performs SHM with a time period T and amplitude a. The magnitude of average velocity of the particle over the time interval during which it travels a distance a/2from the extreme position is :a)a/Tb)2a/Tc)3a/Td)a/2TCorrect answer is option 'C'. Can you explain this answer?
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A particle performs SHM with a time period T and amplitude a. The magnitude of average velocity of the particle over the time interval during which it travels a distance a/2from the extreme position is :a)a/Tb)2a/Tc)3a/Td)a/2TCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle performs SHM with a time period T and amplitude a. The magnitude of average velocity of the particle over the time interval during which it travels a distance a/2from the extreme position is :a)a/Tb)2a/Tc)3a/Td)a/2TCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle performs SHM with a time period T and amplitude a. The magnitude of average velocity of the particle over the time interval during which it travels a distance a/2from the extreme position is :a)a/Tb)2a/Tc)3a/Td)a/2TCorrect answer is option 'C'. Can you explain this answer?.
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