Two equal masses hung from two massless springs of spring constant $k_{1}$ and $k_{2}$ . They have equal maximum velocity when executing simple harmonic motion. The ratio of their amplitudes is:

{(\frac{k_{1}}{k_{2}})}^{1 / 2}

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(\frac{k_{1}}{k_{2}})

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(\frac{k_{2}}{k_{1}})

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{(\frac{k_{2}}{k_{1}})}^{1 / 2}

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(\frac{k_{1}^{2}}{k_{2}^{2}})

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Solution

The correct option is E ${(\frac{k_{2}}{k_{1}})}^{1 / 2}$
As the masses are executing SHM the maximum velocity, $v = a ω$
$∵ v_{1} = v_{2}$ (given)
$∴ a_{1} ω_{1} = a_{2} ω_{2}$
$\frac{a_{1}}{a_{2}} = \frac{ω_{2}}{ω_{1}}$ ..........(i)
We know that the time period,
$T = 2 π \sqrt{\frac{m}{k}}$
$\frac{T}{2 π} = \sqrt{\frac{m}{k}}$
$\Rightarrow \frac{1}{ω} = \sqrt{\frac{m}{k}}$
$\Rightarrow ω = \sqrt{\frac{k}{m}}$
$\Rightarrow ω \propto \sqrt{k}$ ........(ii)
From Eqs. (i) and (ii), we get
$\frac{a_{1}}{a_{2}} = \sqrt{\frac{k_{2}}{k_{1}}}$ .

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