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zorro
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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?