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The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be
Options
(a) T
(b) T / √2
(c) 2T
(d) √2T
Correct Answer:
√2T
Explanation:
T = 2π √(m / K)
T₁ / T₂ = √(M₁ /M₂)
T₂ = T₁√(M₂ / M₁) = T₁ √(2M / M)
T₂ = T₁ √2 = √2 T (where T₁ = T)
Related Questions: - The two ends of a rod of length L and a uniform cross-sectional area
- The breaking stress of a wire depends upon
- The sum of the magnitudes of two forces acting at a point is 16 N
- The instantaneous angular position of a point on a rotating wheel is given by
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The two ends of a rod of length L and a uniform cross-sectional area
- The breaking stress of a wire depends upon
- The sum of the magnitudes of two forces acting at a point is 16 N
- The instantaneous angular position of a point on a rotating wheel is given by
- A point charge q is situated at a distance r on axis from one end of a thin
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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