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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: - In a transistor, the value of βis 100, then the value of α will be
- In the diagram shown in figure, match the following about weight
- A magnetic force acting on a charged particle of charge -2µC in a magnetic field of 2T
- A particle of mass 1 mg has the same wavelength as an electron moving with a velocity
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In a transistor, the value of βis 100, then the value of α will be
- In the diagram shown in figure, match the following about weight
- A magnetic force acting on a charged particle of charge -2µC in a magnetic field of 2T
- A particle of mass 1 mg has the same wavelength as an electron moving with a velocity
- Two identical flutes produce fundamental notes of frequency 300 Hz at 27° C
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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