1) Two pendulums of same amplitude but time period $3s$ and $7s$ start oscillating simultaneously from two opposite extreme positions. After how much time they will be in phase?

2) A ring of mass $m$ and radius $R$ rests in equilibrium on a smooth cone of semi vertical angle $45^{o}$,The radius of the cone is $2R$. the radius of circular cross section of the ring is $r(r<<R)$.

a) What will be the tension in the ring ?

b) What will be the speed of transverse wave on the ring?

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## Comments

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TopNewestAnswer to second problem @Tanishq Varshney

Upward component of normal force balances weight of the chain. The horizontal component causes tension in ring. We can find tension in chain using the value of horizontal component.

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$N$ in the normal force acting on a small part of the ring.

$dm*g$ is the weight of that part and Tension acts in the ring along the tangent.

So $\frac{N}{\sqrt{2}}=dmg ...... (i)$

and $2Tsin(\frac{d \theta}{2}) =\frac{N}{\sqrt{2}} .........(ii)$

Using $(i)$ & $(ii)$ we get,

$2Tsin(\frac{d \theta}{2})=\lambda Rd\theta g$ where $\lambda$ is the linear mass density of the ring.

So $Td \theta=\lambda Rd\theta g$

and $T=\frac { M }{ 2\pi } g$

For finding the speed of the transverse wave , use $v_t = \sqrt{\frac{T}{\lambda}}$

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@Kushal Patankar @Nishant Rai @Rohit Shah @Abhineet Nayyar

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@satvik pandey @Raghav Vaidyanathan

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What are the answers?

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I think the first problem's answer is $21/8 s$

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i want the method

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Equation of motion of first pendulum is: $x_1=-acos(\frac{2\pi} {3}t)$

Equation of motion of the second pendulum is: $x_2=acos(\frac{2\pi} {7}t)$

Just find $t$ for which $x_1=x_2$ and also velocities in same direction.

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answers please @Tanishq Varshney

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for the first one $\frac{21}{8}$ and second a) $\frac{mg}{2\pi}$ b) $\sqrt{gr}$. b) part is easy once a) part is known

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I got second one.

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img

In the figure $N$ in the normal force acting on a small part of the string $(dm)g$ is the weight of that part. And Tension acts in the string along the tangent.

So $N/ \sqrt{2}=dmg$....(1)

and $2Tsin(\frac{d \theta}{2}) =N/ \sqrt{2}$....(2)

Using (1) and (2) we get

$2Tsin(\frac{d \theta}{2})=\lambda Rd\theta g$........................ ($\lambda=\frac { M }{ R2\pi } )$

So $Td \theta=\lambda Rd\theta g$

and $T=\frac { M }{ R2\pi } R$

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$\omega$ what will be the tension in the ring now??

if the cone is rotated about its axis with angular velocityLog in to reply

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The rotation of the ring and rod are independent.

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We can find the maximum angular velocity that the rough cone can have so that the ring doesn't slip on it for a given value of $\mu$.

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$T$ is dimensionally incorrect. Correcr answer = $T = \frac{M}{2\pi}g$

your answer toLog in to reply

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Try this. Tension in the Ring!

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