1) Two pendulums of same amplitude but time period and start oscillating simultaneously from two opposite extreme positions. After how much time they will be in phase?
2) A ring of mass and radius rests in equilibrium on a smooth cone of semi vertical angle ,The radius of the cone is . the radius of circular cross section of the ring is .
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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Answer 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.
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 2N=dmg......(i)
and 2Tsin(2dθ)=2N.........(ii)
Using (i) & (ii) we get,
2Tsin(2dθ)=λRdθg where λ is the linear mass density of the ring.
So Tdθ=λRdθg
and T=2πMg
For finding the speed of the transverse wave , use vt=λT
@Kushal Patankar @Nishant Rai @Rohit Shah @Abhineet Nayyar
@satvik pandey @Raghav Vaidyanathan
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What are the answers?
I think the first problem's answer is 21/8s
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i want the method
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Equation of motion of first pendulum is: x1=−acos(32πt)
Equation of motion of the second pendulum is: x2=acos(72πt)
Just find t for which x1=x2 and also velocities in same direction.
answers please @Tanishq Varshney
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for the first one 821 and second a) 2πmg b) gr. b) part is easy once a) part is known
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I got second one.
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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/2=dmg....(1)
and 2Tsin(2dθ)=N/2....(2)
Using (1) and (2) we get
2Tsin(2dθ)=λRdθg........................ (λ=R2πM)
So Tdθ=λRdθg
and T=R2πMR
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ω 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 μ.
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T is dimensionally incorrect. Correcr answer = T=2πMg
your answer toLog in to reply
Try this. Tension in the Ring!