JEE Main 2014PhysicsOscillationsHardMCQ

JEE Main 2014 — Oscillations Question with Solution

JEE Main 2014 (09 Apr Online)

Question

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).

Choose an option

Show full solutionCorrect option: B
Correct answer
B60 N
 

Step-by-step explanation


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

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About this question

This is a previous-year question from JEE Main 2014, covering the Oscillations chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.