Two springs of equal lengths are equal cross-sectional areas are made of materials whose Young's moduli are in the ratio of $2 : 3$ . They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of?

9 : 4

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3 : 2

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3 \sqrt{3} : 2 \sqrt{2}

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\sqrt{3} : \sqrt{2}

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Solution

The correct option is D $\sqrt{3} : \sqrt{2}$
As we know ,
$T = 2 π \sqrt{\frac{m}{k}}$
now as we know
$k \propto Y$
so,
$T \propto \frac{1}{\sqrt{Y}}$
then,
$\frac{T_{1}}{T_{2}} = \frac{\sqrt{Y_{2}}}{\sqrt{Y_{1}}}$

$\frac{T_{1}}{T_{2}} = \frac{\sqrt{3}}{\sqrt{2}}$
hence option D is correct.

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Two springs of equal lengths are equal cross-sectional areas are made of materials whose Young's moduli are in the ratio of 2:3. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of?

The correct option is D √3:√2As we know ,T=2π√mknow as we know k∝Yso,T∝1√Ythen,T1T2=√Y2√Y1T1T2=√3√2hence option D is correct.

Two springs of equal lengths are equal cross-sectional areas are made of materials whose Young's moduli are in the ratio of $2 : 3$ . They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of?

The correct option is D $\sqrt{3} : \sqrt{2}$
As we know ,
$T = 2 π \sqrt{\frac{m}{k}}$
now as we know
$k \propto Y$
so,
$T \propto \frac{1}{\sqrt{Y}}$
then,
$\frac{T_{1}}{T_{2}} = \frac{\sqrt{Y_{2}}}{\sqrt{Y_{1}}}$

$\frac{T_{1}}{T_{2}} = \frac{\sqrt{3}}{\sqrt{2}}$
hence option D is correct.