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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?
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
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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.
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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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