Two masses m 1 and m 2 are connected by a spring of constant k. The reduced mass is μ = m 1m _2/(m 1 +m 2 ). The angular frequency of oscillation is:

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Two masses m₁ and m₂ are connected by a spring of constant k. The reduced mass is μ = m_1m_2/(m₁+m₂). The angular frequency of oscillation is:

Answer: √(k/μ).

A √(k/μ)
B √(k(m₁+m₂))
C √(k/m₁ + k/m₂)
D √(kμ)

Correct answer: A. √(k/μ)

Explanation: The two-body oscillator reduces to a one-body problem with reduced mass μ. ω = √(k/μ).

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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Two masses m1 and m2 are connected by a spring of constant k. The reduced mass is μ = m1m_2/(m1+m2). The angular frequency of oscillation is:

Concept context

More Oscillations questions

Two masses m₁ and m₂ are connected by a spring of constant k. The reduced mass is μ = m_1m_2/(m₁+m₂). The angular frequency of oscillation is: