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#extreme position

3 public questions tagged with this topic.

Energy in SHM - Kinetic, Potential and Total

In an ideal SHM system, what occurs to the total mechanical energy as the particle moves from the mean position to an ex

**Energy distribution** shows maximum kinetic at equilibrium and maximum potential at extremes, sum constant. This relation enables calculation of amplitude, velocity at any displacement via v = ±√(2(E-U)/m), and understanding of energy storage in oscillating system for NEET problems. Total mechanical energy in ideal SHM (no friction) is conserved, remaining constant as kinetic energy converts to potential energy during the motion. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It remains constant follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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Energy in SHM - Kinetic, Potential and Total

In SHM, if the particle is at an extreme position, what can be inferred about its kinetic energy?

**Energy in SHM** interconverts between kinetic K = ½ m v² = ½ m ω² (A² - x²) and potential U = ½ k x² = ½ m ω² x², total E = K + U = ½ k A² = ½ m ω² A² constant, independent of time. At mean position x=0, E = K_max = ½ m ω² A², at extremes x=±A, E = U_max = ½ k A². At the extreme position ( x = ± A ), velocity is zero ( v = 0 ), so kinetic energy ( K = (1/2) m v² ) is zero, with all energy being

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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Energy in SHM - Kinetic, Potential and Total

In an ideal SHM system, what happens to the kinetic energy as the particle approaches the extreme position?

**Conservation of mechanical energy** in undamped SHM implies total energy proportional to amplitude squared A² and spring constant k. E = ½ k A² allows amplitude determination from known E and k via A = √(2E/k), with k = m ω² linking dynamical and energetic descriptions for spring-mass system. Kinetic energy decreases as the particle approaches the extreme position, where velocity becomes zero, and potential energy reaches its maximum. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It decreases follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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