Two particles A and B of equal masses are suspended from two massless springs of spring constant k_1 and k_2 , respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of A and B is

Correct answer: C. k_2 k_1 V_ =ω A=m ω^2 I =k ω^2= k m ω= k m ω K or (ω_1)/(ω_2)= k_1 k_2 V_ A =V_ B ( k_1 ~m )(A_1)=( k_2 ~m )(A_2) A_1 ~A_2 = k_2 k_1

JEE Main 2003

Exam: JEE Main
Year: 2003
Subject: Physics
Chapter: Oscillations

JEE Main 2003 · Physics · Oscillations

Two particles A and B of equal masses are suspended from two...

Two particles

\mathrm{A}

and

\mathrm{B}

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

\mathrm{k}_1

and

\mathrm{k}_2

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

A

and

B

Options

A $\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_2}}$
B $\frac{\mathrm{k}_2}{\mathrm{k}_1}$
C $\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}}$
D $\frac{\mathrm{k}_1}{\mathrm{k}_2}$

Correct answer

C. $\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}}$

Step-by-step solution

\mathrm{V}_{\max }=\omega

\mathrm{A}=\mathrm{m}

\omega^2

\mathrm{I}

=\mathrm{k}

\Rightarrow \omega^2=\frac{\mathrm{k}}{\mathrm{m}} \Rightarrow \omega=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \quad \Rightarrow \omega \propto \sqrt{\mathrm{K}} \quad

\quad \frac{\omega_1}{\omega_2}=\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_2}}

\mathrm{V}_{\mathrm{A}} \max =\mathrm{V}_{\mathrm{B}} \max \quad \Rightarrow\left(\sqrt{\frac{\mathrm{k}_1}{\mathrm{~m}}}\right)\left(\mathrm{A}_1\right)=\left(\sqrt{\frac{\mathrm{k}_2}{\mathrm{~m}}}\right)\left(\mathrm{A}_2\right) \Rightarrow \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}}

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