A mass in the diagram below undergoes simple harmonic When the mass reaches point x = 0 its instantaneous acceleration is:

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In simple harmonic motion, the acceleration a is proportional to the displacement x and directed toward the equilibrium position, given by a = -ω^2 x.
Option 1: Zero. Since x = 0 at the equilibrium position, the displacement is zero, and thus a = -ω^2(0) = 0. So the instantaneous acceleration at x = 0 vanishes, regardless of the velocity.
Option 2: Less than maximum. The magnitude of acceleration is maximum at the maximum displacements ±A, where |x| = A, and zero at x = 0. Therefore saying it is less than maximum at x = 0 is incorrect because it is exactly zero, not some nonzero value.
Option 3: Constant and doesn’t depend on the location. This would imply a is the same for all x, which contradicts a = -ω^2 x, where acceleration depends linearly on the displacement. Hence this statement is false.
Option 4: Maximum and can be positive or negative. The maximum magnitude occurs at the turning points x = ±A, not at x = 0. Also the sign of acceleration is determined by the direction toward the origin; at x = 0 the acceleration is zero, not maximum or having a definite sign.

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A mass in the diagram below undergoes simple harmonic When the mass reaches point x = 0 its instantaneous acceleration is:

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