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  2. I have a doubt, kindly clarify. - Oscillations and Waves - JEE Main
# Engineering

Two particles A and B of equal masses are suspended from two massless springs of spring constants k_{1} and k_{2} respectively. If the maximum velocities, during oscillations, are equal, the ratio of amplitudes of A and B is

  • Option 1)

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

  • Option 2)

    \; \; \; \; k_{2}/k_{1}\;

  • Option 3)

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

  • Option 4)

    \; k_{1}/k_{2}

 

Answers (1)
P prateek

As we learnt in

Relation of velocity and displacement -

v= w\sqrt{A^{2}-x^{2}}
 

- wherein

\rightarrow  x is displacement from mean position

\rightarrow  A is Amplitude.

 

 

 

Maximum velocity under simple harmonic motion v_{m}=a\omega

\therefore \; \; \; v_{m}=\frac{2\pi a}{T}=(2\pi a)\left ( \frac{1}{T} \right )=(2\pi a)\left ( \frac{1}{2\pi }\sqrt{\frac{k}{m}} \right )

or\; \; \; v_{m}=a\sqrt{\frac{k}{m}}

\because \; \; \; (v_{m})_{A}=(v_{m})_{B}

\therefore \; \; \; a_{1}\sqrt{\frac{k_{1}}{m}}=a_{2}\sqrt{\frac{k_{2}}{m}}\Rightarrow \frac{a_{1}}{a_{2}}=\sqrt{\frac{k_{2}}{k_{1}}}

Correct option is 3.


Option 1)

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

This is an incorrect option.

Option 2)

\; \; \; \; k_{2}/k_{1}\;

This is an incorrect option.

Option 3)

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

This is the correct option.

Option 4)

\; k_{1}/k_{2}

This is an incorrect option.

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