A question

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

Note by Tanishq Varshney
3 years, 9 months ago

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1 vote

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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, v0=2ghF=mgv_0 = \sqrt{2gh}\\F = mg

Assuming χ\chi to be the stiffness of the spring.

Writing equation of position of the mass-plank system w.r.t initial position, let ω2=χM+m\omega^2 = \dfrac \chi{M+m} (M+m)y¨=mgχyy¨=mM+mgω2y(M+m)\ddot y = mg - \chi y\\\ddot y = \dfrac m{M+m}g - \omega^2y

Solving this differential equation, y=mgχmgχcosωt+v0ωsinωty=\dfrac{mg}\chi - \dfrac{mg}\chi \cos\omega t+\dfrac {v_0} \omega \sin \omega t

Time period, T=2πω=2πM+mχT = \dfrac{2\pi}\omega = 2\pi\sqrt{\dfrac{M+m}\chi}

Now, either using conservation of energy or taking maximum value of yy in the above equations, h=b2χ2bgm2gmh = \frac{b^2 \chi-2 b g m}{2 g m} OR h=a2χ2+2agχm+2agχMg2M22gχmh = \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 hh are correct and that I did not miss out any terms in my equation... ¨\ddot\smile

Hope this helps!

Kishore S Shenoy - 3 years, 9 months ago

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

f=12π2mg(ba)(M+m)f=\frac{1}{2 \pi}\sqrt{\frac{2mg}{(b-a)(M+m)}} and

h=M+mm.abbah=\frac{M+m}{m}.\frac{ab}{b-a}

Tanishq Varshney - 3 years, 9 months ago

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But what is wrong in my method?

Kishore S Shenoy - 3 years, 9 months ago

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@Kishore S Shenoy I know, I took the value of χ\chi to be given...

Kishore S Shenoy - 3 years, 9 months ago

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@Kishore S Shenoy Could u look àt my solution in which I have tagged you and tell the flaw.

Tanishq Varshney - 3 years, 9 months ago

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You can use a+b2+mgχ=b\dfrac {a+b} 2+\dfrac {mg} \chi =b Solving,T=2π(ba)(M+m)2mg T=2\pi \sqrt{\dfrac {(b-a)(M+m)} {2mg}}

Kishore S Shenoy - 3 years, 9 months ago

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@Kishore S Shenoy How did u get this expression , plz explain.

Tanishq Varshney - 3 years, 9 months ago

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

Kishore S Shenoy - 3 years, 9 months ago

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

At the equilibrium position, y¨=0\ddot y=0

So, yeq=mgχ, downwardsy_{eq} = \dfrac{mg}\chi,\text{ downwards}

Also, a+b2=A, the amplitude\dfrac{a+b}2 = A,\text{ the amplitude}

So yeq+b=A=a+b2mgχ=ba2χ=2mgbay_{eq} + b = A = \dfrac{a+b }2\\\Rightarrow \dfrac{mg}\chi = \dfrac{b-a}2\\\chi = \dfrac{2mg}{b-a}

As we know, T=2πm+Mχ=2π(ba)(m+M)2mgT = 2\pi \sqrt{\dfrac{m+M}\chi}=2\pi\sqrt{\dfrac{(b-a)(m+M)}{2mg}}

Now, to find hh,Mg=ky0Mg = ky_0mg(h+b)+Mgb=k2(y0+b)2k2y02=k2(b2+2by0)mg(h+b)=k2b2\begin{aligned}mg(h+b) + Mgb &= \dfrac k2(y_0+b)^2-\dfrac k2 y_0^2\\ &=\dfrac k 2 (b^2+2by_0)\end{aligned}\\ \boxed{mg(h+b) = \dfrac k 2 b^2} mg(ha)Mga=k2(ay0)2k2y02=k2(a22ay0)mg(ha)=k2a2\begin{aligned}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{aligned} \\ \boxed{mg(h-a)=\dfrac k 2 a^2} Dividing, h+bha=b2a2h+b=b2bah=abba\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 - 3 years, 9 months ago

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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.

Kishore S Shenoy - 3 years, 9 months ago

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

Tanishq Varshney - 3 years, 9 months ago

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

Kishore S Shenoy - 3 years, 9 months ago

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@Kishore S Shenoy Ok may be the answer i have is incorrect. So your answer is correct :)

Tanishq Varshney - 3 years, 9 months ago

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I'll post this in you question?

Kishore S Shenoy - 3 years, 9 months ago

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