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A point performs simple harmonic oscillation of period T and the equation of motion is given by x = a sin (ωt + π/6). After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?                                                                                                                                                       [2008]

  • a)
    T/8

  • b)
    T/6 

  • c)
    T/3

  • d)
    T/12

Correct answer is option 'D'. Can you explain this answer?
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Understanding the Problem:
Given equation of motion: x = a sin (ωt + π/6)
We need to find the fraction of time period after which the velocity of the point will be equal to half of its maximum velocity.

Calculating Velocity:
The velocity of the point can be calculated by differentiating the equation of motion with respect to time.
v = dx/dt = aω cos(ωt + π/6)

Calculating Maximum Velocity:
To find the maximum velocity, we need to consider the magnitude of the velocity at any point in the oscillation, which is given by |v|max = aω

Calculating the Time:
To find the time at which the velocity is half of its maximum value, we equate the velocity to half of the maximum velocity and solve for time.
0.5aω = aω cos(ωt + π/6)
Solving the above equation gives:
cos(ωt + π/6) = 0.5
ωt + π/6 = ± π/3
ωt = ± π/3 - π/6
Solving for the positive sign gives:
ωt = π/6
t = T/12
Therefore, after T/12 time period, the velocity of the point will be equal to half of its maximum velocity, which corresponds to option 'D'.
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A point performs simple harmonic oscillation of period T and the equation of motion is given by x = a sin (ωt + π/6). After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity? [2008]a)T/8b)T/6c)T/3d)T/12Correct answer is option 'D'. Can you explain this answer?
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