1.
A light spring AB, having natural length a and modulus of elastocoty 3amn^2, lies straight and at its natural length at rest on a horizontal table. A particle of mass m is attached to the end A.
The end B is then moved in a straight line in the direction AB with constant speed V. The resulting motion of the particle is resisted by a force of magnitude 4mnv where v is the speed of the particle. If x is the extension of the spring at time t , show that
d^2x/dt^2+4ndx/dt+3n^2x=4nV
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totaljj
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- 30-06-2004 13:03
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- 30-06-2004 21:12
B is moving at V m/s
A is moving at v m/s
Therefore extension is x = (V-v) m every sec,
i.e. dx/dt = (V-v) m/s
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Tension in spring is T where
T = λx/l
T = (3amn²).x/a where λ = 3amn² and l = a
T = 3mn²x
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Accelerating force on particle at A is given by T-F where F = 4mnv,
T-F = mdv/dt
3mn²x - 4mnv = mdv/dt
3n²x - 4nv = dv/dt
But dx/dt = V-v
or
v = V - dx/dt
dv/dt = -d²x/dt²
substituting for v and dv/dt,
3n²x - 4n(V - dx/dt) = -d²x/dt²
d²x/dt² + 4n.dx/dt + 3n²x = 4nV
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totaljj
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- 30-06-2004 21:43
thank you
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