1) For the system shown in the figure the acceleration of the mass \(m_{4}\) immediately after the lower thread \(x\) is cut will be (assume the threads and springs are light and there is no friction) The diagram is above mass \(m_{4}\) is with string and above it is \(m_{3}\) towards left of \(m_{3}\) is \(m_{1}\).

2) If \(f(x)\) is an increasing function and \(\large{\displaystyle \int_{0}^{x} 2xf^{2}(t).dt=(\displaystyle \int_{0}^{x} 2f(x-t).dt)^{2}}\) for \(f(1)=1\) and \(f(x)\) is continuous for \(x>0\). Find \(f(x)\)

3) Let \(f(x)=sin^{4} \pi x\) and \(g(x)=lnx\). Find the whole area bounded by \(y=f(x)\) and \(y=g(x)\) and \(x=0\)

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TopNewestNice problems.

Just write free body diagrams for each block. What you need to find is the force in the right hand side spring. Once the string is cut, the force in the right hand side spring is the only thing other than gravity acting on \(m_4\). After solving using F.B.D, you will get the force.

Firstly, substitute \(x-t\) in place of \(t\) in the RHS integral. Then take derivative w.r.t \(x\) on both sides. Use the given equation and substitute RHS in place of LHS wherever necessary after taking derivative. If done right, you will now be having an equation in which you have \(\int _{ 0 }^{ x }{ f\left( x \right) dx } \), \(f(x)\) and \(x\) only. You can now write this as a differential equation where \(y=\int _{ 0 }^{ x }{ f\left( x \right) dx }\). Since the differential equation is of 2nd degree, you will have to use quadratic formula. Finally, after solving, you will get two solutions, but only one of them satisfies all conditions given in problem.

\(f(x)\) and \(g(x)\) intersect only once and that is at \(x=1\). Hence the required answer is \(\int _{ 0 }^{ 1 }{( \sin ^{ 4 }{ \pi x } +\log { x }) }dx = \frac{11}{8}\)

@Tanishq Varshney

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Hi Raghav! Could you please help me here

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but in the 3) they have five solutions. why we took only \(0\) to \(1\)

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That is correct, but there the areas are not also bounded by \(x=0\).

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Answers i have are 1) \([\frac{m_{3}+m_{4}-m_{1}-m_{2}}{m_{4}}]g\) \(\quad\) 2) \(f(x)=x^{\sqrt{2}+1}\) \(\quad\) 3) \(\frac{11}{8}\)

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Let the tension in the above string be \(T\). And let the extension of left and right springs be \(x\) and \(x_{2}\).

From the free body diagrams we get

\(T=m_{1}g+kx\).....................(1)

\(kx=m_{2}g\)...............................(2)

\(m_{3}g+kx_{2}=T\)............................(3)

From 1 and 2 we get \(T=(m_{1}+m_{2})g\)

From (2)

\(kx_{2}=(m_{1}+m_{2}-m_{3})g\)

Now when the lower string is cut then only gravitational force and force spring due to string will act on it.

Let the acceleration of \(m_{4}\) be \(a\) be in downward direction---

So \(m_{4}g-kx_{2}=m_{4}a\)

So \(a=\frac { g(m_{ 4 }+m_{ 3 }-m_{ 1 }-m_{ 2 }) }{ m_{ 4 } } \)

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Thanx for the solution \(\ddot \smile\)

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You can solved it orally , by using the newton's 2nd law , directly : Since Tension's is internal forces ...

\({ F }_{ ext }=ma\)

system is moving with same acceleration , ideal strings , inextinsible , so

\({ F }_{ ext }=ma\\ { m }_{ 3 }g+{ m }_{ 4 }g-{ m }_{ 1 }g-{ m }_{ 2 }g={ m }_{ 4 }a\)

hence the answer....

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@Raghav Vaidyanathan @satvik pandey @Nishant Rai @rachit parikh

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