Tough Problem in S.H.M.

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  • #1
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Homework Statement



A particle of mass m is attached to one end of a light elastic string of natural length a and force constant mg/a. The other end of the string is attached to a fixed point X. If the particle is released from rest at X, find time that elapses before it returns to X.


The Attempt at a Solution



Refer the figure made by me (not given in question)
I got the time taken from X to P and P to X [2√2a/g]

The problem I am facing is the time taken to travel from P to B and back
I got the equation of SHM as
x=Asin(ωt + ∂) where A = √3a and ω = √(g/a)
at t = 0 let x = a so that sin∂ = 1/√3
The particle will be again at P after time (say T)
a = √3a sin(ωT + ∂)
sin∂ = sin(ωT + ∂)
∂ = π - (ωT + ∂)
T = 1/ω ( π - 2arcsin( 1/√3 ) )
the total time taken is t = 2√2a/g + √a/g(π - 2arcsin( 1/√3 ))

The answer is t = 2√2a/g + 2√a/g(Π - arccos(1/√3))

In the solution, he took the equation of S.H.M. in terms of cos (not sin) and he too took
t = 0 at x=a
How can the two times taken be different no matter what equation you take???
 

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Answers and Replies

  • #2
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Your phase at t=0 is arcsin(1/√3) and at equilibrium it is 0 but phase can not decrease wth time!
Actually at t=0, x=-a. You can also see why they took the cos function-it is compatible with the increasing phase.
It would be much more convenient if you were to assume x=0 at your supposed x=a.
 
  • #3
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Your phase at t=0 is arcsin(1/√3) and at equilibrium it is 0 but phase can not decrease wth time!

Thanks for your reply.
I did not understand why phase can not decrease with time. Do you mean to say that phase is constant for a SHM?
 
  • #4
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Phase is ωt+Φ---- ω is positive and time t increases with time:biggrin:.
Φ is your phase constant.
 
  • #5
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But arcsin(1/√3) is phase constant not phase......
and phase constant is ∂ in my equation
 
  • #6
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At t=0 arcsin(1/√3) is phase as well as phase constant. But when x=0 which happens at t>0 phase and not phase constant is zero.
 
  • #7
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One more thing- Till where does the particle perform SHM during upward motion? its amplitude is √3 a
But we calculate its time period only till a distance 'a' from the mean position and not √3a.
In the answer,
t = 2√2a/g + 2√a/g(Π - arccos(1/√3))
The term on the left is obtained considering the acceleration to be 'g' during the motion from X to P and P to X. But in reality the particle executes SHM till a point A which is in between X and P. So acc. to me it is under the effect of gravity alone from A to X (not in SHM) and we have to find the time period of SHM from mean position to A and not P.
 
  • #8
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But in reality the particle executes SHM till a point A which is in between X and P.
No as soon as string acquires its natural length it tension becomes zero and particle is under free fall----- that is the qualitative difference b/w a string and a spring.
 
  • #9
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Thanks a ton!! It cleared all the paradox!!
 

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