The time period of oscillation of a spring-mass system in a satellite orbiting Earth is: MCQ with Answer and Explanation

The time period of oscillation of a spring-mass system in a satellite orbiting Earth is:

A. Zero

B. Infinite

C. Depends on orbital radius

D. Same as on Earth

Answer: Option D

Solution (By JKExamLibrary)

Spring-mass period T = 2π√(m/k), depending only on mass and spring constant, not on gravity. In orbit, apparent weightlessness doesn't affect spring force (which depends on displacement, not gravity). Thus period remains unchanged. Pendulum period would change (depends on g), but spring-mass does not. Memory aid: 'Spring period: independent of g; pendulum period: depends on g'. This conceptual question tests oscillations fundamentals, frequently examined in competitive exams. Always distinguish gravity-dependent (pendulum) from gravity-independent (spring-mass) oscillators.

This question belongs to: Science Physics

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Practice More Physics Questions

Question #1 Report Error

A body floats in a liquid with one-third of its volume above the surface. The density of the body relative to the liquid is:

A. 3/2

B. 2/3

C. 1/3

D. 3

View Answer & Solution Discuss

Correct Answer: Option B

Explanation:
By Archimedes' principle, weight of body = weight of displaced liquid. Let V be total volume, ρ_b body density, ρ_l liquid density. Volume submerged = (2/3)V. Thus ρ_b·V·g = ρ_l·(2V/3)·g ⇒ ρ_b/ρ_l = 2/3. Fraction submerged equals density ratio. Memory tip: 'Fraction submerged = ρ_object / ρ_fluid'. This direct application of floatation condition is common in competitive exams. Always identify submerged fraction correctly: here 'one-third above' means two-thirds submerged. Such problems test conceptual clarity in buoyancy applications.

This question belongs to: Science Physics