Verified Solution • 2025–26 Session moderate

QJEEM21Ph2da7: JEE Main 2021 Physics - Class 11 JEEM21Ph2da7 Solution: Two particles A and B of equal masses are suspended from two…

Two particles A and B of equal masses are suspended from two massless springs of spring constants K1 and K2 respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is

Multiple Choice Options

  • A K1K2{{{K_1}} \over {{K_2}}}
  • B K1K2\sqrt {{{{K_1}} \over {{K_2}}}}
  • C K2K1{{{K_2}} \over {{K_1}}}
  • D K2K1\sqrt {{{{K_2}} \over {{K_1}}}}

Step-by-Step Verified Solution

∵\because Vmax⁡=Aω{V_{\max }} = A\omega

Given, ω1A1=ω2A2{\omega _1}{A_1} = {\omega _2}{A_2}

We know that ω=Km\omega = \sqrt {{K \over m}}

∴\therefore k1mA1=k2mA2\sqrt {{{{k_1}} \over m}} {A_1} = \sqrt {{{{k_2}} \over m}} {A_2}

⇒\Rightarrow A1A2=k2k1{{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over {{k_1}}}}

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