Question

The figure shows a particle of mass $' m',$ attached with four identical springs, each of length $l .$ Initial tension in each spring is $F_{0} .$ Neglecting gravity, calculate period of small oscillations of the particle along a line perpendicular to the plane of the figure.

Accepted Answer

Let the particle be displaced slightly through a displacement $d x$ along a line normal to the plane of the figure by applying a $d F$ so that the particle is still in equilibrium. Since, springs are identical, therefore increase in tension of each spring will be the same. Let this increase be $d F_{0} .$ Let at this time, the spring force in each spring be $F_{0} + d F_{0} .$ The horizontal components of this force will get cancelled and we will be left with vertical upward components of four springs.

For the equilibrium of the particle, $d F = 4 (F_{0} + d F_{0}) \sin θ \dots (i)$

As $θ$ is small angle we can write $\sin θ ≃ \tan θ .$ Now Eq. (i) can be written as

$d F = 4 (F_{0} + d F_{0}) \tan θ = 4 (F_{0} + d F_{0}) \frac{d x}{l} = 4 F_{0} \frac{d x}{l} + 4 d F_{0} \frac{d x}{l}$

Neglecting product of small quantities above equation can be written as $d F = \frac{4 F_{0}}{l} \cdot d x .$ Hence $\frac{d F}{d x} = \frac{4 F_{0}}{l} = K_{eff} .$

It means $ω^{2} = \frac{K_{eff}}{m} = \frac{4 F_{0}}{m l}$ or $T = 2 π \sqrt{\frac{m l}{4 F_{0}}} = π \sqrt{\frac{m l}{F_{0}}}$

B M Sharma Solutions for Chapter: Linear and Angular Simple Harmonic Motion, Exercise 1: ILLUSTRATIONS

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Linear and Angular Simple Harmonic Motion, Exercise 1: ILLUSTRATIONS

Questions from B M Sharma Solutions for Chapter: Linear and Angular Simple Harmonic Motion, Exercise 1: ILLUSTRATIONS with Hints & Solutions

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