Two bodies M and N of equal masses are suspended from two separate mass less springs of spring constants k_1 and k_2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitudes of M to that of N is

  • Option 1)

    \frac{k_1}{k_2}

  • Option 2)

    \sqrt{\frac{k_1}{k_2}}

  • Option 3)

    \frac{k_2}{k_1}

  • Option 4)

    \sqrt{\frac{k_2}{k_1}}

 

Answers (1)

As 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.

 

 V_{max.} = Aw = A\sqrt{\frac{K}{m}}                                                 \left [ \because \frac{K}{m} = w^{2} \right ]

V_{1} = A\sqrt{\frac{K_{1}}{m}}

V_{2} = A\sqrt{\frac{K_{2}}{m}}

\therefore Ratio = \frac{V_{2}}{V_{1}} = \sqrt{\frac{K_{2}}{K_{1}}}


Option 1)

\frac{k_1}{k_2}

This option is incorrect

Option 2)

\sqrt{\frac{k_1}{k_2}}

This option is incorrect

Option 3)

\frac{k_2}{k_1}

This option is incorrect

Option 4)

\sqrt{\frac{k_2}{k_1}}

This option is correct

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