| ⇦ | 
| ⇨ |
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 which of the following systems will the radius of the first orbit (n=1) be minimum?
- If force is proportional to square of velocity, then the dimension of proportionality
- A body of length 1 m having cross-sectional area 0.75 m² has heat flow through it
- A spring has time period T. It is cut into n equal parts. The time period of each part
- To get three images of single object, one should have two plane mirrors at an angle of
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In which of the following systems will the radius of the first orbit (n=1) be minimum?
- If force is proportional to square of velocity, then the dimension of proportionality
- A body of length 1 m having cross-sectional area 0.75 m² has heat flow through it
- A spring has time period T. It is cut into n equal parts. The time period of each part
- To get three images of single object, one should have two plane mirrors at an angle of
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