Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25 rad s-1, and amplitude 1.6 cm while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is (take g = 10 m s-2).
PhysicsOscillationsJEE MainJEE Main 2014 (09 Apr Online)
Options:
  • A 20 N
     
  • B 60 N
     
  • C 40 N
     
  • D 10 N
     
Solution:
2024 Upvotes Verified Answer
The correct answer is: 60 N
 


Given ω = 2 5 rad/s and A = 1.6 cm

∴   ω=km=25 k= 25 2

Maximum compression in the spring

=xo+A

=mgk+A

  Fmax=4g + k(x0 + A)

Fmax =4g+kmgk+A

            = 40 + 20 = 60 N

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.