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⚛️ Physics  ·  Oscillations  ·  NEET & JEE

A particle in SHM of amplitude 10 cm has a speed of 8 cm/s when its displacement is 6 cm. Its time period is:

Answer: 6.28 s.

  • A 3.14 s
  • B 1.57 s
  • C 12.56 s
  • D 6.28 s

Correct answer: D. 6.28 s

Explanation: v = &omega;&radic;(A<sup>2</sup> - x<sup>2</sup>) gives 8 = &omega;&radic;(100 - 36) = 8&omega;, so &omega; = 1 rad/s. Then T = 2&pi;/&omega; = 6.28 s.

Displacement, Velocity, Acceleration in SHMx(t)v(t)a(t)v leads x by 90°; a is exactly out of phase with x (a = −ω²x)

In SHM, displacement and velocity are 90° out of phase (v is maximum when x=0, and zero when x is extreme), while acceleration is always exactly opposite in sign to displacement, consistent with a = −ω²x.

Concept context

Simple harmonic motion, spring-mass, pendulum, energy in SHM, and resonance.

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