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JEE Main 2014 (9 Apr, Shift 1) · Spring Systems

Spring Systems question from JEE Main 2014 (9 Apr, Shift 1)

Q88JEE Main 2014 · 9 APR · SHIFT 1Difficult

Spring Systems

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, 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 ms ______2).

Figure

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Answer: (c)

  1. Maximum spring compression is at the lowest point of the 1 kg mass: spring force =mg+mω2A= mg + m\omega^2 A.
  2. mg+mω2A=1×10+1×252×0.016=10+10=20 Nmg + m\omega^2 A = 1 \times 10 + 1 \times 25^2 \times 0.016 = 10 + 10 = 20\text{ N}
  3. Force on floor =Mg+20=40+20=60 N= Mg + 20 = 40 + 20 = 60\text{ N}
  4. So the answer is (c) 60 N.
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