A particle mass 0.1 kg is held between two rigid supports by two springs of force constant $8 N m^{- 1}$ and $2 N m^{- 1}$ . If the particle is displaced along the direction of the length of the springs, its frequency of vibration is

\frac{5}{π}

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\frac{8}{π}

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\frac{2}{π}

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\frac{1}{π}

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Solution

The correct option is A $\frac{5}{π}$ Hz
Given,
$m = 0.1 k g$
$K_{1} = 8 N / m$
$K_{2} = 2 N / m$
From the above figure,
Springs are in parallel
$K_{e q} = K_{1} + K_{2}$
$K_{e q} = 8 + 2 = 10 N / m$
The frequency of the oscillation,
$f = \frac{1}{2 π} \sqrt{\frac{K_{e q}}{m}}$

$f = \frac{1}{2 π} \sqrt{\frac{10}{0.1}} = \frac{10}{2 π}$

$f = \frac{5}{π} H z$
The correct option is A.

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A particle mass 0.1 kg is held between two rigid supports by two springs of force constant 8Nm−1 and 2Nm−1. If the particle is displaced along the direction of the length of the springs, its frequency of vibration is

The correct option is A 5πHzGiven,m=0.1kgK1=8N/mK2=2N/mFrom the above figure,Springs are in parallelKeq=K1+K2Keq=8+2=10N/mThe frequency of the oscillation,f=12π√Keqmf=12π√100.1=102πf=5πHzThe correct option is A.

A particle mass 0.1 kg is held between two rigid supports by two springs of force constant $8 N m^{- 1}$ and $2 N m^{- 1}$ . If the particle is displaced along the direction of the length of the springs, its frequency of vibration is