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Mechanics
22 Feb 2007 11:59:38 IST
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elasticity
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one end of a long metallic wire of length l is tied to da ceiling . the other end is tied to a massless spring of spring const k.a masss hangs freely frm free end of da spring.the area of cs and the youngs modullus of elasticity of da wire are a&y resp.if mass slightly pulled down &released it ll osscilatewid time period t=?
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Manasi
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22 Feb 2007 12:15:34 IST
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the timeperiod will be 2pi rt(l/g)....
y would it change??? when mass is hanged, it will bring some extension in length of the rod, but its time period will not change, as the mass is freely hanging, its equilibrium position will change, but its time-period will remain same.....
i think m right, do tell me!!!
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22 Feb 2007 17:29:40 IST
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i am explaing this problem with a very easy concept.....
1st take a smaller problem in which mass is attached to only wire and thr is no spring. whn the mass is pulled down it will execute SHM due to the elastic nature of the metallic wire.
sol:.....for this whn mass is pulled down a small distance "x".
the forces acting on it are T(by wire) and Mg due to gravity....restoring force = Ma (a is acceleration upwards).....
thus , Ma = T - Mg
or a = T/M - g .......(1)
Now we have Y = (T/A) / (x/L) .......(Y=stress/starin)
=> T = AYx/L ....... (2)
from 1 and 2 ====>> a = (AY/ML)x - g
thus a (dx2/dt2) is proprtional to x ...condition of SHM
hence time period = 2pi root (ML/AY) ........(3)
1st part complete.......
for a general spring time period = 2pi root (M/K) ..........(4)
comparing 3 and 4 we hav K = AY/L .......thus we can consider the wire as the spring of spring constant K which is equal to AY/L.....
CURRENT problem is tht we hav two springs in series with spring contants k and AY/L...
1/Knet = 1/k + 1/(AY/L) = 1/k + L/AY
thus u have Knet and time period is thus 2pi root ( M / Knet)
Ne querries??????
1st take a smaller problem in which mass is attached to only wire and thr is no spring. whn the mass is pulled down it will execute SHM due to the elastic nature of the metallic wire.
sol:.....for this whn mass is pulled down a small distance "x".
the forces acting on it are T(by wire) and Mg due to gravity....restoring force = Ma (a is acceleration upwards).....
thus , Ma = T - Mg
or a = T/M - g .......(1)
Now we have Y = (T/A) / (x/L) .......(Y=stress/starin)
=> T = AYx/L ....... (2)
from 1 and 2 ====>> a = (AY/ML)x - g
thus a (dx2/dt2) is proprtional to x ...condition of SHM
hence time period = 2pi root (ML/AY) ........(3)
1st part complete.......
for a general spring time period = 2pi root (M/K) ..........(4)
comparing 3 and 4 we hav K = AY/L .......thus we can consider the wire as the spring of spring constant K which is equal to AY/L.....
CURRENT problem is tht we hav two springs in series with spring contants k and AY/L...
1/Knet = 1/k + 1/(AY/L) = 1/k + L/AY
thus u have Knet and time period is thus 2pi root ( M / Knet)
Ne querries??????
23 Feb 2007 01:50:07 IST
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When a weight is added to a spring the spring elongates by the following equation;
Y = mgL/(lA)
this force is acting upwards if the mass is pulled a little down then the force is = YA(l+x)/L
the restoring force. This equation can be compared to F=kx on comparison it gives k=YA/L. Since there is also spring attached to the spring therefore the springs are in series and are added following
1/K_comb = 1/k + 1/K
on combination the spring constant is YAK/(YA+KL)
T=2pisqrt(M/ ((YAK/(YA+KL)))
---Hope this helps Prabhat
Give me your rate
Y = mgL/(lA)
this force is acting upwards if the mass is pulled a little down then the force is = YA(l+x)/L
the restoring force. This equation can be compared to F=kx on comparison it gives k=YA/L. Since there is also spring attached to the spring therefore the springs are in series and are added following
1/K_comb = 1/k + 1/K
on combination the spring constant is YAK/(YA+KL)
T=2pisqrt(M/ ((YAK/(YA+KL)))
---Hope this helps Prabhat
Give me your rate
, wire will behave as a pring.... well the solution is already thr
25 Feb 2007 04:02:14 IST
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Any chemical bond results from the accumulation of charge density in the binding region to an extent sufficient to balance the forces of repulsion. Ionic and covalent binding represent the two possible extremes of reaching this state of electrostatic equilibrium and there is a complete spectrum of bond densities lying between these two extremes. Since covalent and ionic charge distributions exhibit radically different chemical and physical properties, it is important, if we are to understand and predict the bulk properties of matter, to know which of the two extremes of binding a given molecule most closely approximates.
The dipole moment is defined as the product of the total amount of positive or negative charge and the distance between their centroids
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The dipole moment is defined as the product of the total amount of positive or negative charge and the distance between their centroids
Best Wishes
25 Feb 2007 04:56:53 IST
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Solution :The wire would also act as an spring whose spring constant will be YA/L.
There are two springs in the series.
Hence the equivalent spring constant is given by
(1/knet) = (1/kwire) + (1/kspring)
= (L/YA) + (1/kspring)
Time period = 2
m/knet Best Wishes
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