Oscillations

JEE Main level, numerical answer, medium

85 of 194

Question 85 of 194

A mass 0.9kg0.9kg, attached to a horizontal spring, executes SHM with an amplitude A1{A}_{1}. When this mass passes through its mean position, then a smaller mass of 124g124g is placed over it and both masses move together with amplitude A2{A}_{2}. If the ratio A1A2\dfrac{{A}_{1}}{{A}_{2}} is αα−1\dfrac{\alpha }{\alpha -1}, then the value of α\alpha will be _____ .

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Answer: 16

At mean position, the velocity of the particle in S.H.M. is maximum, so kinetic energy is KE=12kA2KE=\dfrac{1}{2}k{A}^{2}, where, kk is spring constant.
We also know that, kinetic energy is related to momentum as KE=p22mKE=\dfrac{{p}^{2}}{2m}.
Now, from both equations, we have
12kA2=p22m\dfrac{1}{2}k{A}^{2}=\dfrac{{p}^{2}}{2m} (Here, momentum is conserved)
⇒(A1A2)2=m2m1=1024900\Rightarrow {\left(\dfrac{{A}_{1}}{{A}_{2}}\right)}^{2}=\dfrac{{m}_{2}}{{m}_{1}}=\dfrac{1024}{900}
⇒A1A2=3230=1615=1616−1\Rightarrow \dfrac{{A}_{1}}{{A}_{2}}=\dfrac{32}{30}=\dfrac{16}{15}=\dfrac{16}{16-1}
∴α=16∴\alpha =16

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