⇦ | ⇨ |
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 maximum kinetic energy of the photoelectrons depends only on
- Two bodies A and B having temperature 327° C and 427° C are radiating heat
- In the given figure, a diode D is connected to an external resistance R=100 Ω
- The two ends of a rod of length L and a uniform cross-sectional area
- A tuned amplifier circuit is used to generate a carrier frequency of 2 MHz
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The maximum kinetic energy of the photoelectrons depends only on
- Two bodies A and B having temperature 327° C and 427° C are radiating heat
- In the given figure, a diode D is connected to an external resistance R=100 Ω
- The two ends of a rod of length L and a uniform cross-sectional area
- A tuned amplifier circuit is used to generate a carrier frequency of 2 MHz
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply