| ⇦ |
| ⇨ |
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: - A body takes 5 minute for cooling from 50⁰C to 40⁰C. Its temperature
- The point of suspension λ of a simple pendulum with normal time period T₁ is moving
- A thin circular ring of mass M and radius R is rotating in a horizontal plane about
- The magnetic flux linked with a circuit of resistance 100Ω increases from 10 to 60 Wb.
- A solid body rotates about a stationary axis, so that its angular velocity
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body takes 5 minute for cooling from 50⁰C to 40⁰C. Its temperature
- The point of suspension λ of a simple pendulum with normal time period T₁ is moving
- A thin circular ring of mass M and radius R is rotating in a horizontal plane about
- The magnetic flux linked with a circuit of resistance 100Ω increases from 10 to 60 Wb.
- A solid body rotates about a stationary axis, so that its angular velocity
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