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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Simple Harmonic Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Simple Harmonic Motion

Three springs A , B , and C , each having a force constant k , are...

Three springs AA, BB, and CC, each having a force constant kk, are attached to a particle of mass mm. Initially, the angle between any two springs is 120^{}^{{}^{\circ}}. If the particle is slightly pushed against spring CC and then released, determine its time period of oscillation.

Options

  1. A2π2m3k2\pi \sqrt{\dfrac{2m}{3k}}
  2. B2πmk2\pi \sqrt{\dfrac{m}{k}}
  3. C2π3m2k2\pi \sqrt{\dfrac{3m}{2k}}
  4. D2πm3k2\pi \sqrt{\dfrac{m}{3k}}

Correct answer

A. 2π2m3k2\pi \sqrt{\dfrac{2m}{3k}}

Step-by-step solution

Let the displacement of the particle be xx towards spring CC.

The compression in spring CC is xx. The restoring force exerted by spring CC is FC=kxF_C = kx, directed opposite to the displacement.

The angle between the line of action of spring CC and those of springs AA and BB is 120^{}^{{}^{\circ}}.

The extension in springs AA and BB is the component of displacement along their respective lengths, which is x \cos(180^{}^{{}^{\circ}} - 120^{}^{{}^{\circ}}) = x \cos 60^{}^{{}^{\circ}} = \dfrac{x}{2}.

The restoring force exerted by each of springs AA and BB is FA=FB=k(x2),directedtowardstheirrespectivefixedends.<br><br>ThenetrestoringforcefromspringsF_A = F_B = k \left(\dfrac{x}{2}\right), directed towards their respective fixed ends.<br><br>The net restoring force from springsAandandBalongthedirectionoppositetothedisplacementis:<br><br>along the direction opposite to the displacement is:<br><br>F_{A,B} = F_A \cos 60^{}^{{}^{\circ}} + F_B \cos 60^{}^{{}^{\circ}} = 2 \left(\dfrac{kx}{2}\right) \left(\dfrac{1}{2}\right) = \dfrac{kx}{2}

The total restoring force on the particle is:

Fnet=FC+FA,B=kx+F_{net} = F_C + F_{A,B} = kx +\dfrac{kx}{2} = \dfrac{3kx}{2}

The effective spring constant of the system is keff=k_{eff} =\dfrac{3k}{2}.

The time period of oscillation is given by:

T=2πT = 2\pi\sqrt{\dfrac{m}{k_{eff}}} ==2\pi \sqrt{\dfrac{2m}{3k}}
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