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JEE Main 2017 (8 Apr, Shift 1) · Spring Systems

Spring Systems question from JEE Main 2017 (8 Apr, Shift 1)

Q85JEE Main 2017 · 8 APR · SHIFT 1Medium

Spring Systems

A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is:

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Answer: (c)

  1. f=12πkmf = \frac{1}{2\pi}\sqrt{\frac{k}{m}}: 1=12πk11 = \frac{1}{2\pi}\sqrt{\frac{k}{1}}.
  2. New system: k′=2kk' = 2k, m′=8m' = 8 kg, so f′=12π2k8=12⋅12πkf' = \frac{1}{2\pi}\sqrt{\frac{2k}{8}} = \frac{1}{2}\cdot\frac{1}{2\pi}\sqrt{k}.
  3. f′=12f' = \frac{1}{2} Hz, so the answer is (c).
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