Q69 / 90 · JEE Main 2014 (09 Apr Online) · Physics

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).
Figure

  • A
    20 N
     
  • Correct answer
    B
    60 N
     
  • C
    40 N
     
  • D
    10 N
     

Solution

Figure
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

Hence, the correct answer is option: 'B'.

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