A mass \(M\) is in static equilibrium on a massless vertical spring as shown. A ball of mass m dropped from certain height sticks to the mass \(M\) after colliding with it. The oscillations they perform reach to height 'a' above original level of scales and depth \(b\) below it.

a) Find time period of oscillations.

b) What is the height above the initial level from which the mass m was dropped?

Please post your detailed solution.

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## Comments

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TopNewest@Kushal Patankar @Akhil Bansal

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@Kishore S Shenoy @Abhishek Sharma

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WARNING : I don't believe this solution to be flawless...

At the time when the ball just sticks to the plank, given an inelastic collision, \[v_0 = \sqrt{2gh}\\F = mg\]

Assuming \(\chi\) to be the stiffness of the spring.

Writing equation of position of the mass-plank system w.r.t initial position, let \(\omega^2 = \dfrac \chi{M+m}\) \[(M+m)\ddot y = mg - \chi y\\\ddot y = \dfrac m{M+m}g - \omega^2y\]

Solving this differential equation, \[y=\dfrac{mg}\chi - \dfrac{mg}\chi \cos\omega t+\dfrac {v_0} \omega \sin \omega t\]

Time period, \[T = \dfrac{2\pi}\omega = 2\pi\sqrt{\dfrac{M+m}\chi}\]

Now, either using conservation of energy or taking maximum value of \(y\) in the above equations, \[h = \frac{b^2 \chi-2 b g m}{2 g m}\] OR \[h = \frac{a^2 \chi^2+2 a g \chi m+2 a g\chi M-g^2 M^2}{2 g \chi m}\]

I hope the final values, of \(h\) are correct and that I did not miss out any terms in my equation... \(\ddot\smile\)

Hope this helps!

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First of all thank you for replying , but the answer I have

\(f=\frac{1}{2 \pi}\sqrt{\frac{2mg}{(b-a)(M+m)}}\) and

\(h=\frac{M+m}{m}.\frac{ab}{b-a}\)

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But what is wrong in my method?

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You can use \[\dfrac {a+b} 2+\dfrac {mg} \chi =b\] Solving,\[ T=2\pi \sqrt{\dfrac {(b-a)(M+m)} {2mg}} \]

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@Tanishq Varshney I get \[h = \dfrac{ab}{b-a}\]

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At the time when the body sticks to the pan of the spring, \[\ddot y = \dfrac m{m+M}g - \dfrac\chi {m+M}y\] Here, downwards is taken to be positive.

At the equilibrium position, \[\ddot y=0\]

So, \[y_{eq} = \dfrac{mg}\chi,\text{ downwards}\]

Also, \[\dfrac{a+b}2 = A,\text{ the amplitude}\]

So \[y_{eq} + b = A = \dfrac{a+b }2\\\Rightarrow \dfrac{mg}\chi = \dfrac{b-a}2\\\chi = \dfrac{2mg}{b-a}\]

As we know, \[T = 2\pi \sqrt{\dfrac{m+M}\chi}=2\pi\sqrt{\dfrac{(b-a)(m+M)}{2mg}}\]

Now, to find \(h\),\[Mg = ky_0\]\[\begin{align*}mg(h+b) + Mgb &= \dfrac k2(y_0+b)^2-\dfrac k2 y_0^2\\ &=\dfrac k 2 (b^2+2by_0)\end{align*}\\ \boxed{mg(h+b) = \dfrac k 2 b^2}\] \[\begin{align*}mg(h-a)-Mga &= \dfrac k 2 (a-y_0)^2-\dfrac k 2 y_0^2\\&=\dfrac k 2 (a^2-2ay_0)\end{align*} \\ \boxed{mg(h-a)=\dfrac k 2 a^2}\] Dividing, \[\dfrac{h+b}{h-a} = \dfrac{b^2}{a^2} \\h+b = \dfrac{b^2}{b-a}\\h = \dfrac{ab}{b-a}\]

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@Tanishq Varshney , I hope this helps! I'm quite sure about these values... please point out if there is any mistake.

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yup it will , thank you once again , but I feel the answer for h is not correct.

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But I see no mistake in my solution.. do you see any? (Also, my friend got the same value of \(h\) when I gave this problem to him today... that means the answer should match... :(

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I'll post this in you question?

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