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

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Restating the scenario: A mass in a simple harmonic oscillator moves along x and reaches the extreme position x = +A. The question asks for its instantaneous acceleration at that moment.

Option 1: Zero. This would be true if the velocity were zero and the net forces canceled, but in SHM the acceleration is proportional to displacement and points toward the equilibrium, so at x = +A the acceleration is not zero.

Option 2: Maximum and positive. The sign here would imply acceleration pointing in the positive x direction. In SHM, the acceleration is always directed toward the equilibrium (opposite to the displacement). At x = +A, the displacement is positive, so the acceleration must be negative, not positive.

Option 3: Constant and doesn’t depend on the location. Acceleration in SHM is not constant; it depends on the instantaneous displacement x through a = -ω^2 x. Therefore this statement is incorrect.

Option 4: Maximum and negative. This aligns with a = -ω^2 x. At x = +A, the magnitude of acceleration is ω^2 A (maximum possible) and its direction is negative (toward the equilibrium at x = 0). Hence this option correctly describes both the magnitude and the sign.

In summary, the acceleration in a simple harmonic oscillator is proportional to the negative of the displacement, so at the furthest right point x = +A the acceleration is at its largest magnitude and directed left (negative). This makes the fourth option the correct one.

SPHY001 Practice Quiz - Unit 16 | LA056

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

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