Discussion Forum



1)

Two bodies  A and B of equal mass are suspended from two separate massless springs of force constant k1 and k2, respectively. The bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitudes  of body  A to that of body B is 


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

B) $\frac{k_{2}}{k_{1}}$

C) $\frac{k_{1}}{k_{2}}$

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

Answer:

Option A

Explanation:

 The maximum velocity of body in SHM

                                   v= A $\omega$

 where , A is amplitude  of body and $\omega$  is the angular frequency.

 It is given that , the maximum velocities of bodies are equal

                       $V_{A}=V_{B}$

 $A_{A}\omega_{A}=A_{B}\omega_{B}$

 $A_{A}\sqrt{\frac{k_{1}}{m}}=A_{B}\sqrt{\frac{k_{2}}{m}}$           

                                                                                    $\left(\because \omega =\sqrt{\frac{k}{m}}\right)$

 $\frac{A_{A}}{A_{B}}=\sqrt{\frac{k_{2}}{k_{1}}}$