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QJEEM17Ph3b00: JEE Main 2017 Physics - Class 11 JEEM17Ph3b00 Solution: A 1 kg block attached to a spring vibrates with a frequency of…

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 :

Multiple Choice Options

  • A 14Hz{1 \over 4}Hz
  • B 122Hz{1 \over {2\sqrt 2 }}Hz
  • C 12Hz{1 \over 2}Hz
  • D 22 HzHz

Step-by-Step Verified Solution

For 1 kg block :

JEE Main 2017 (Online) 8th April Morning Slot Physics - Simple Harmonic Motion Question 123 English Explanation 1

Here frequency of spring (f) = 12πkm{1 \over {2\pi }}\sqrt {{k \over m}}

Given that, F = 1 Hz

∴   \therefore\,\,\, 12πk1{1 \over {2\pi }}\sqrt {{k \over 1}} = 1

⇒\Rightarrow    \,\,\, k = 4 π\pi 2 N m −- 1

For 8 kg Block :

JEE Main 2017 (Online) 8th April Morning Slot Physics - Simple Harmonic Motion Question 123 English Explanation 2

Here two identical springs are attached in parallel. So,

Keq = k + k = 2k

∴   \therefore\,\,\, Frequency of 8 kg block,

F' = 12πkeqm′{1 \over {2\pi }}\sqrt {{{{k_{eq}}} \over {m'}}}

= 12π2k8{1 \over {2\pi }}\sqrt {{{2k} \over 8}}

= 12π2×4π28{1 \over {2\pi }}\sqrt {{{2 \times 4{\pi ^2}} \over 8}}

= 12π×π{1 \over {2\pi }} \times \pi

= 12{1 \over 2} Hz

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