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⚛️ Physics  ·  Oscillations  ·  NEET & JEE

Two springs of constants 100 N/m and 200 N/m are joined in series and carry a 3 kg mass. The time period of SHM is about:

Answer: 1.33 s.

  • A 0.67 s
  • B 1.00 s
  • C 1.33 s
  • D 2.00 s

Correct answer: C. 1.33 s

Explanation: Series combination: k<sub>eff</sub> = (100 &times; 200)/(100 + 200) = 66.7 N/m. T = 2&pi;&radic;(m/k<sub>eff</sub>) = 2&pi;&radic;(3/66.7) = 2&pi;(0.212) = 1.33 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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