Two springs of constants k1 and k2 equal maximum velocities, when executing simple harmonic motion. The ratio of their amplitudes (masses are equal) will be
 

  • Option 1)

    \frac{k_1}{k_2}

  • Option 2)

    \frac{k_2}{k_1}

  • Option 3)

    (\frac{k_1}{k_2})^{\frac{1}{2}}

  • Option 4)

    (\frac{k_2}{k_1})^{\frac{1}{2}}

 

Answers (1)
V Vakul

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 = Aw = A\sqrt{\frac{k}{m}}

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

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

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


Option 1)

\frac{k_1}{k_2}

This option is incorrect

Option 2)

\frac{k_2}{k_1}

This option is incorrect

Option 3)

(\frac{k_1}{k_2})^{\frac{1}{2}}

This option is incorrect

Option 4)

(\frac{k_2}{k_1})^{\frac{1}{2}}

This option is correct

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