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JEE Main 20236 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
6 April
Shift
Morning Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2023 · 6 April · Morning Shift · Physics · Laws of Motion

A mass m is attached to two springs as shown in figure. The spring...

A mass mm is attached to two springs as shown in figure. The spring constants of two springs are K1{K}_{1} and K2.{K}_{2} . For the frictionless surface, the time period of oscillation of mass mm is

A mass m is attached to two springs as shown in figure. The spring constants of two springs are K _ 1 and K _ 2 . For

Options

  1. A2πmK1+K22\pi\sqrt{\dfrac{m}{{K}_{1} + {K}_{2}}}
  2. B12πK1−K2m\dfrac{1}{2\pi}\sqrt{\dfrac{{K}_{1} - {K}_{2}}{m}}
  3. C2πmK1−K22\pi\sqrt{\dfrac{m}{{K}_{1} - {K}_{2}}}
  4. D12πK1+K2m\dfrac{1}{2\pi}\sqrt{\dfrac{{K}_{1} + {K}_{2}}{m}}

Correct answer

A. 2πmK1+K22\pi\sqrt{\dfrac{m}{{K}_{1} + {K}_{2}}}

Step-by-step solution

A mass m is attached to two springs as shown in figure. The spring constants of two springs are K _ 1 and K _ 2 . For

For the given combination, if the block is displaced towards right, elongation in one spring will be equal to the compression in other and the direction of force due to both spring will be the same and restoring in nature. Hence, both springs are effectively in parallel combination.

Therefore,

⇒Keff=K1+K2⇒ {K}_{eff} = {K}_{1} + {K}_{2} and

⇒T=2πmKeff=2πmK1+K2⇒ T = 2\pi\sqrt{\dfrac{m}{{K}_{eff}}} = 2\pi\sqrt{\dfrac{m}{{K}_{1} + {K}_{2}}}

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