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

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In simple harmonic motion, the velocity is given by v = ±ω√(A^2 − x^2), where A is the amplitude and x is the displacement from equilibrium. At the equilibrium position x = 0, the expression becomes v = ±ωA, which has the largest magnitude possible for the velocity in this motion. The sign (positive or negative) indicates the direction of motion along the x-axis.

Option 1: Zero. This would imply the mass is momentarily at rest at x = 0, which is false for SHM because the restoring force is maximum there and the mass passes through x = 0 with maximum speed.

Option 2: Constant and doesn’t depend on the location. In SHM the velocity depends on how far you are from equilibrium; it is largest at x = 0 and decreases as |x| increases toward ±A. Thus this statement is incorrect.

Option 3: Less than maximum. The velocity is not merely less than the maximum at x = 0; at x = 0 it reaches its maximum magnitude. This option misstates that the speed is smaller than its peak when at equilibrium.

Option 4: Maximum and can be positive or negative. This aligns with the SHM relation v = ±ωA at x = 0, indicating the velocity is at its maximum magnitude and its sign reflects the direction of motion (positive or negative).

SPHY001 Practice Quiz - Unit 16 | LA056

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

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