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NEET PHYSICSOscillationsMedium

Question

A body of mass mm is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass mm is slightly pulled down and released, it oscillates with a time period of 3 s3\text{ s}. When the mass mm is increased by 1 kg1\text{ kg}, the time period of oscillations becomes 5 s5\text{ s}. The value of mm in kg is:

A

3/4

B

4/3

C

16/9

D

9/16

Step-by-Step Solution

The time period (TT) of a spring-mass system undergoing simple harmonic motion is given by the formula T=2πmkT = 2\pi \sqrt{\frac{m}{k}}, where mm is the mass and kk is the spring constant.

  1. Case 1: For mass mm, the time period is 3 s3\text{ s}. 3=2πmk3 = 2\pi \sqrt{\frac{m}{k}}

  2. Case 2: When the mass is increased by 1 kg1\text{ kg}, the new mass is (m+1)(m + 1) and the time period becomes 5 s5\text{ s}. 5=2πm+1k5 = 2\pi \sqrt{\frac{m+1}{k}}

  3. Divide the two equations: 35=2πm/k2π(m+1)/k\frac{3}{5} = \frac{2\pi \sqrt{m/k}}{2\pi \sqrt{(m+1)/k}} 35=mm+1\frac{3}{5} = \sqrt{\frac{m}{m+1}}

  4. Solve for mm: Squaring both sides: 925=mm+1\frac{9}{25} = \frac{m}{m+1} 9(m+1)=25m9(m + 1) = 25m 9m+9=25m9m + 9 = 25m 16m=916m = 9 m=916 kgm = \frac{9}{16} \text{ kg}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Oscillations. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSOscillationsattachedspringspringnegligibleslightly

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