Two spring mass systems have equal masses and springs of spring constants $k_{1} and k_{2}$ . If the maximum velocities in the two systems are equal, then the ratio of amplitude of the first system to that of the second is :

\sqrt{\frac{k_{1}}{k_{2}}}

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\frac{k_{1}}{k_{2}}

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\frac{k_{2}}{k_{1}}

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\sqrt{\frac{k_{2}}{k_{1}}}

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Solution

The correct option is D $\sqrt{\frac{k_{2}}{k_{1}}}$
In spring mass system, the maximum velocity is,
$v_{m a x} = A ω$
Here, $ω = \sqrt{\frac{k}{m}}$
$∴ v_{m a x} = A \sqrt{\frac{k}{m}}$

$\Rightarrow \frac{v_{m a x_{1}}}{v_{m a x_{2}}} = \frac{A_{1} \sqrt{\frac{k_{1}}{m}}}{A_{2} \sqrt{\frac{k_{2}}{m}}} = \frac{A_{1} \sqrt{k_{1}}}{A_{2} \sqrt{k_{2}}}$
As, $v_{m a x_{1}} = v_{m a x_{2}}$

$\Rightarrow \frac{A_{1}}{A_{2}} = \sqrt{\frac{k_{2}}{k_{1}}}$

Hence, $(D)$ is the correct answer.

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Two spring mass systems have equal masses and springs of spring constants k1 and k2. If the maximum velocities in the two systems are equal, then the ratio of amplitude of the first system to that of the second is :

The correct option is D √k2k1In spring mass system, the maximum velocity is, vmax=Aω Here, ω=√km ∴vmax=A√km ⇒vmax1vmax2=A1 √k1mA2 √k2m=A1 √k1A2 √k2 As, vmax1=vmax2 ⇒A1A2=√k2k1 Hence, (D) is the correct answer.

Two spring mass systems have equal masses and springs of spring constants $k_{1} and k_{2}$ . If the maximum velocities in the two systems are equal, then the ratio of amplitude of the first system to that of the second is :