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Q.Derive an expression for the velocity of a particle in simple harmonic motion and show its graph of variation with time.

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2024Subjective· 3mImportance★★★★★
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Figure — Stem says 'show its graph of variation with time' for SHM velocity; the catalog figure shows displacement, vel
Figure — Stem says 'show its graph of variation with time' for SHM velocity; the catalog figure shows displacement, vel

For x = A sin(ωt + φ), differentiating gives v = Aω cos(ωt + φ) = ω√(A^2 − x^2); v is maximum at the mean position and zero at the extremes, and its time graph is a cosine curve.

Let the displacement of a particle in SHM be:

x = A sin(ωt + φ)

where A is amplitude, ω is angular frequency, and φ is the initial phase. Velocity is the time derivative of displacement:

v = dx/dt = Aω cos(ωt + φ)

To express v in terms of x (rather than t), use sin^2 θ + cos^2 θ = 1. Since sin(ωt+φ) = x/A, we get cos(ωt+φ) = √(1 − x^2/A^2). So:

v = Aω √(1 − x^2/A^2) = ω√(A^2 − x^2)

Behaviour:

  • At the mean position, x = 0: v = ω√(A^2 − 0) = Aω — this is the MAXIMUM speed (v_max = Aω).
  • At the extreme positions, x = ±A: v = ω√(A^2 − A^2) = 0 — velocity is ZERO there (momentarily at rest before reversing). …

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