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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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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