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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TopNewestAt 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, \[\dfrac{h+b}{h-a} = \dfrac{b^2}{a^2} \\h+b = \dfrac{b^2}{b-a}\\h = \dfrac{ab}{b-a}\] – Kishore S Shenoy · 1 year ago

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– Tanishq Varshney · 1 year ago

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

– Kishore S Shenoy · 1 year ago

I'll post this in you question?Log in to reply

– Kishore S Shenoy · 1 year ago

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... :(Log in to reply

– Tanishq Varshney · 1 year ago

Ok may be the answer i have is incorrect. So your answer is correct :)Log in to reply

@Tanishq Varshney , I hope this helps! I'm quite sure about these values... please point out if there is any mistake. – Kishore S Shenoy · 1 year ago

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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! – Kishore S Shenoy · 1 year ago

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

\(h=\frac{M+m}{m}.\frac{ab}{b-a}\) – Tanishq Varshney · 1 year ago

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– Kishore S Shenoy · 1 year ago

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

– Tanishq Varshney · 1 year ago

How did u get this expression , plz explain.Log in to reply

@Tanishq Varshney I get \[h = \dfrac{ab}{b-a}\] – Kishore S Shenoy · 1 year ago

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– Kishore S Shenoy · 1 year ago

But what is wrong in my method?Log in to reply

– Kishore S Shenoy · 1 year ago

I know, I took the value of \(\chi\) to be given...Log in to reply

– Tanishq Varshney · 1 year ago

Could u look àt my solution in which I have tagged you and tell the flaw.Log in to reply

@Kishore S Shenoy @Abhishek Sharma – Tanishq Varshney · 1 year ago

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@Kushal Patankar @Akhil Bansal – Tanishq Varshney · 1 year ago

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