Final answer:
The correct displacement of a particle in SHM when its speed is half the maximum is √3/2 A. This is because the potential energy at this point is three-fourths of the maximum potential energy, which correlates to the square of the displacement from the mean position.
option a . √3/2 A is correct.
Step-by-step explanation:
The question relates to a particle performing linear simple harmonic motion (SHM) with an amplitude A and a time period T. The velocity at half the maximum speed corresponds to half the maximum kinetic energy, which can be represented by v(t) = -Umax sin (wt + φ), where Umax is the maximum speed. The displacement x at the time when the velocity is half the maximum speed can be found using the expression for velocity in SHM and the conservation of energy principle.
From the given energy relationship for a simple harmonic oscillator, when the velocity is Umax/2, the kinetic energy is one-fourth the maximum kinetic energy, since kinetic energy is proportional to the square of velocity. Therefore, the potential energy at this point is three-fourths the maximum potential energy. If we consider that the total energy of the system is given by the sum of kinetic and potential energy, and it is constant throughout the motion, we can assert that at this point, the potential energy must be three-fourths of the total energy. Hence, x should be A times the square root of 3 divided by 2, because potential energy in a SHM system is proportional to the square of the displacement from the mean position.
Therefore, the correct answer is that the particle's displacement when the speed is half the maximum is √3/2 A.