Wanted help in these

1) For the system shown in the figure the acceleration of the mass m4m_{4} immediately after the lower thread xx is cut will be (assume the threads and springs are light and there is no friction) The diagram is above mass m4m_{4} is with string and above it is m3m_{3} towards left of m3m_{3} is m1m_{1}.

2) If f(x)f(x) is an increasing function and 0x2xf2(t).dt=(0x2f(xt).dt)2\large{\displaystyle \int_{0}^{x} 2xf^{2}(t).dt=(\displaystyle \int_{0}^{x} 2f(x-t).dt)^{2}} for f(1)=1f(1)=1 and f(x)f(x) is continuous for x>0x>0. Find f(x)f(x)

3) Let f(x)=sin4πxf(x)=sin^{4} \pi x and g(x)=lnxg(x)=lnx. Find the whole area bounded by y=f(x)y=f(x) and y=g(x)y=g(x) and x=0x=0

Note by Tanishq Varshney
4 years, 5 months ago

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Nice problems.

  1. 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 m4m_4. After solving using F.B.D, you will get the force.

  2. Firstly, substitute xtx-t in place of tt in the RHS integral. Then take derivative w.r.t xx 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 0xf(x)dx\int _{ 0 }^{ x }{ f\left( x \right) dx } , f(x)f(x) and xx only. You can now write this as a differential equation where y=0xf(x)dxy=\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.

  3. f(x)f(x) and g(x)g(x) intersect only once and that is at x=1x=1. Hence the required answer is 01(sin4πx+logx)dx=118\int _{ 0 }^{ 1 }{( \sin ^{ 4 }{ \pi x } +\log { x }) }dx = \frac{11}{8}

@Tanishq Varshney

Raghav Vaidyanathan - 4 years, 5 months ago

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but in the 3) they have five solutions. why we took only 00 to 11

Tanishq Varshney - 4 years, 5 months ago

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

Raghav Vaidyanathan - 4 years, 5 months ago

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@Raghav Vaidyanathan ok, now i understood, can u please elaborate 2) one

Tanishq Varshney - 4 years, 5 months ago

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Hi Raghav! Could you please help me here

satvik pandey - 4 years, 5 months ago

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Answers i have are 1) [m3+m4m1m2m4]g[\frac{m_{3}+m_{4}-m_{1}-m_{2}}{m_{4}}]g \quad 2) f(x)=x2+1f(x)=x^{\sqrt{2}+1} \quad 3) 118\frac{11}{8}

Tanishq Varshney - 4 years, 5 months ago

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Let the tension in the above string be TT. And let the extension of left and right springs be xx and x2x_{2}.

From the free body diagrams we get

T=m1g+kxT=m_{1}g+kx.....................(1)

kx=m2gkx=m_{2}g...............................(2)

m3g+kx2=Tm_{3}g+kx_{2}=T............................(3)

From 1 and 2 we get T=(m1+m2)gT=(m_{1}+m_{2})g

From (2)

kx2=(m1+m2m3)gkx_{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 m4m_{4} be aa be in downward direction---

So m4gkx2=m4am_{4}g-kx_{2}=m_{4}a

So a=g(m4+m3m1m2)m4a=\frac { g(m_{ 4 }+m_{ 3 }-m_{ 1 }-m_{ 2 }) }{ m_{ 4 } }

satvik pandey - 4 years, 5 months ago

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Thanx for the solution ¨\ddot \smile

Tanishq Varshney - 4 years, 5 months ago

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@Tanishq Varshney You are welcome!! :)

satvik pandey - 4 years, 5 months ago

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

Fext=ma{ F }_{ ext }=ma

system is moving with same acceleration , ideal strings , inextinsible , so
Fext=mam3g+m4gm1gm2g=m4a{ F }_{ ext }=ma\\ { m }_{ 3 }g+{ m }_{ 4 }g-{ m }_{ 1 }g-{ m }_{ 2 }g={ m }_{ 4 }a

hence the answer....

ankit kumar - 4 years, 5 months ago

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