A mass on a spring undergoes SHM. When the mass is at its maximum displacement from equilibrium, its instantaneous velocity

Question

BlackTom AI · Accepted Answer

For simple harmonic motion with displacement x(t) = A cos(ωt + φ), the velocity is v(t) = -Aω sin(ωt + φ). At maximum displacement x = ±A the cosine equals ±1, so the sine term must be zero and therefore v = 0.

1) Relevant concepts
- SHM can be represented by x(t) = A cos(ωt + φ) (or a sine form).  
- Velocity is the time derivative: v(t) = dx/dt = -Aω sin(ωt + φ).  
- Maximum displacement (amplitude) occurs when x = ±A.  
- Energy in SHM oscillates between kinetic and potential; at extreme displacement kinetic energy is zero and potential energy is maximum.

2) Step-by-step reasoning
- Set x(t) = A cos(ωt + φ). Maximum displacement means x = ±A, which happens when cos(ωt + φ) = ±1.  
- At those times cos(θ) = ±1 implies sin(θ) = 0 (where θ = ωt + φ).  
- Substitute into velocity: v(t) = -Aω sin(θ) = -Aω(0) = 0.

3) Intermediate steps (explicit)
- x = ±A → cos(θ)=±1 → θ = nπ (integer n).  
- sin(nπ) = 0 → v = -Aω · 0 = 0.

4) Why the correct answer is correct
- The mathematical form shows velocity must be zero at extremes. Physically, the mass reverses direction at the turning point, so instantaneous speed is zero there while acceleration (and restoring force) is maximal.

5) Why other choices are incorrect
- "is maximum": incorrect because velocity is zero at turning points, not maximal (maximum speed occurs at equilibrium x=0).  
- "is less than maximum, but not zero": incorrect because it specifically is zero, not merely reduced.  
- "cannot be determined": incorrect because SHM analytic form gives a definite value.

Conclusion: The instantaneous velocity at maximum displacement is zero, confirming the correct answer: "is zero."

A mass on a spring undergoes SHM. When the mass is at its maximum displacement from equilibrium, its instantaneous velocitySingle choice

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