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Oscillations Part 1

Section: Physics

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Question : 18 of 100

Marks: +1, -0

Two particles

A

B

of equal masses are suspended from two massless springs of spring constants

k_{1}

k_{2}

, respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of

A

B

[17 Mar 2021 Shift 2]

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

Solution:

We know that, the expression of maximum velocity during oscillation,

V_{\max}=A \omega

Given,

V_{\max}(A)=V_{\max}(B) \Rightarrow A_{1} \omega_{1}=A_{2} \omega_{2}

Now, the ratio of the amplitude during oscillation,

\frac{A_{1}}{A_{2}} = \frac{\omega_{2}}{\omega_{1}}

We know that,

\omega = \sqrt{\frac{k}{m}}

where,

k

is the spring constant,

m

is the mass of the object.

Substituting the value of

\omega

in Eq. (i), we get

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

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