Waste less time on Facebook — follow Brilliant.
×

A question

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.

Note by Tanishq Varshney
11 months, 1 week ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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, \[\dfrac{h+b}{h-a} = \dfrac{b^2}{a^2} \\h+b = \dfrac{b^2}{b-a}\\h = \dfrac{ab}{b-a}\] Kishore S Shenoy · 11 months, 1 week ago

Log in to reply

@Kishore S Shenoy yup it will , thank you once again , but I feel the answer for h is not correct. Tanishq Varshney · 11 months, 1 week ago

Log in to reply

@Tanishq Varshney I'll post this in you question? Kishore S Shenoy · 11 months, 1 week ago

Log in to reply

@Tanishq Varshney 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... :( Kishore S Shenoy · 11 months, 1 week ago

Log in to reply

@Kishore S Shenoy Ok may be the answer i have is incorrect. So your answer is correct :) Tanishq Varshney · 11 months, 1 week ago

Log in to reply

@Kishore S Shenoy @Tanishq Varshney , I hope this helps! I'm quite sure about these values... please point out if there is any mistake. Kishore S Shenoy · 11 months, 1 week ago

Log in to reply

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 · 11 months, 1 week ago

Log in to reply

@Kishore S Shenoy 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

\(h=\frac{M+m}{m}.\frac{ab}{b-a}\) Tanishq Varshney · 11 months, 1 week ago

Log in to reply

@Tanishq Varshney You can use \[\dfrac {a+b} 2+\dfrac {mg} \chi =b\] Solving,\[ T=2\pi \sqrt{\dfrac {(b-a)(M+m)} {2mg}} \] Kishore S Shenoy · 11 months, 1 week ago

Log in to reply

@Kishore S Shenoy How did u get this expression , plz explain. Tanishq Varshney · 11 months, 1 week ago

Log in to reply

@Tanishq Varshney @Tanishq Varshney I get \[h = \dfrac{ab}{b-a}\] Kishore S Shenoy · 11 months, 1 week ago

Log in to reply

@Tanishq Varshney But what is wrong in my method? Kishore S Shenoy · 11 months, 1 week ago

Log in to reply

@Kishore S Shenoy I know, I took the value of \(\chi\) to be given... Kishore S Shenoy · 11 months, 1 week ago

Log in to reply

@Kishore S Shenoy Could u look àt my solution in which I have tagged you and tell the flaw. Tanishq Varshney · 11 months, 1 week ago

Log in to reply

Log in to reply

@Kushal Patankar @Akhil Bansal Tanishq Varshney · 11 months, 1 week ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...