If two springs of spring constants $k_{1}$ and $k_{2}$ while executing SHM have equal highest velocities, then the ratio of their amplitudes will be (their masses are in ratio $1 : 2$ ).

\sqrt{2 k_{2} / k_{1}}

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\sqrt{2 k_{1} / k_{2}}

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2 k_{1} / k_{2}

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2 k_{2} / k_{1}

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Solution

The correct option is A $\sqrt{2 k_{2} / k_{1}}$
Given, $m_{2} = 2 m_{1}$
For a loaded spring, the angular frequency
$ω = \sqrt{k / m}$
$∴ \frac{ω_{1}}{ω_{2}} = \sqrt{\frac{k_{1}}{k_{2}} \times \frac{m_{2}}{m_{1}}} = \sqrt{\frac{k_{1}}{k_{2}} \times 2}$
For equal maximum velocity, $A_{1} ω_{1} = A_{2} ω_{2}$
We have, $\frac{A_{1}}{A_{2}} = \frac{ω_{2}}{ω_{1}} = \sqrt{\frac{k_{2}}{k_{1}} \times 2} = \sqrt{\frac{2 k_{2}}{k_{1}}}$ .

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If two springs of spring constants k1 and k2 while executing SHM have equal highest velocities, then the ratio of their amplitudes will be (their masses are in ratio 1:2).

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If two springs of spring constants $k_{1}$ and $k_{2}$ while executing SHM have equal highest velocities, then the ratio of their amplitudes will be (their masses are in ratio $1 : 2$ ).