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x= Asin(wt)------------------------the particle moves with this trajectory. the wave travels in the y direction.

the question asks us to derive an expression for the distance moved by the particle in time "t" starting fom t=0.

i came up with a solution. but i'm getting stuck at one point. My prof. tells me that higher calculus is rquired for this question which is beyond the scope of the JEE. i think i have an alternate approach. PLease help.

Consider a particle undergoing UCM. describing a circle of radius "A" with angular freq. "w", at any arbitrary position , say (Acosθ , Asinθ), now the projection of the radial vector to this point on the x axis represents the distance of the particle undergoing SHM according to the given eqn. from the origin, i.e. the mean position.

thus, x= Acosθ----------- (1)

=>θ =nπ/2 - wt ---------------(here "n" is an odd integer)

now the arc length the UCM particle covers (lets call it "P" for convinience) corresponds to the linear displacement of the SHM particle (lets call it Q).....now this will be valid only when both these quantities are infinitismly small so we have to use derivatives.

consider an arc length covered by P on sweeping an angle of Δθ, say

L = AΔθ

=> dL/dt = A dθ/dt

= A w

the problem is i am not able to relate the motion of P with that of Q, mathematically. the theory part is pretty clear i suppose.

x= Asin(wt)------------------------the particle moves with this trajectory. the wave travels in the y direction.

the question asks us to derive an expression for the distance moved by the particle in time "t" starting fom t=0.

i came up with a solution. but i'm getting stuck at one point. My prof. tells me that higher calculus is rquired for this question which is beyond the scope of the JEE ( he was saying something about "legendres polynomials"). i think i have an alternate approach. PLease help.

Consider a particle undergoing UCM. describing a circle of radius "A" with aangular freq. "w", at any arbitrary position , say (Acos $\theta$ , Asin $\theta$ ), now the projection of the radial vector to this point on the x axis represents the distance of the particle undergoing SHM according to the given eqn. from the origin, i.e. the mean position.

thus, x= Acos $\theta$ ----------- (1)

=> $\theta$ =n $\pi$ /2 - wt ---------------(here "n" is an odd integer)

now the arc length the UCM particle covers (lets call it "P" for convinience) corresponds to the linear displacement of the SHM particle (lets call it Q).....now this will be valid only when both these quantities are infinitismly small so we have to use derivatives.

consider an arc length covered by P on sweeping an angle of $\Delta$ $\theta$ , say

L = A $\Delta$ $\theta$

=> dL/dt = A d $\theta$ /dt

= A w

the problem is i am not able to relate the motion of P with that of Q, mathematically. the theory part is pretty clear i suppose.

nevermind.. i found my book..thanks for the trouble though..

guys can you provide me a list of genral integral forms..i lost my book and need to revise urgently...please provide me with the list ASAP!!!

Please explain the meaning of eq. wt, no. of equivalents and the relations of both with number of moles, molecular wt. and all other prameters aailable.

Why are Transverse waves slower than longitudial waves?

oh yeah, i forgot...please rate my answer..

well said sid.shah.90..

but i wouldn't go as a far as to say that a point has not been defined after more than 500 years of scientific progress..

in my opinion," a point is a dimensionless quantity in three dimensional space."

i am of course, speaking in purely geometrical hypothesis. we have forgotten the dude's secondary question..

is a point equivalnt to a circle with zero radius? i.e. a point circle???? by my defn., NO! because when you do so, you explicitly express the point in the form

x²+y²=0 or some similar form..

that is, we have used the x and y dimensions...the arguments i read above are pretty much correct especially the one given by sid. BUT a correction would be that "points" as such do not need dimensional definition, which is a definition in itself.

Hence, points are defined.

Q)Find the centre and radius of the smaller of the 2 circles touching the parabola 75y²=64(5x-3) at (6/5,8/5) and the x-axis.

i have identified one key idea that the normal to the parabola at the known pt. will pass through the centre of the required circle. Thus, i have 1 eqn. and 2 variables. The second eqn. is obtained by slope form of the line joining the centre of the circle and the vertex of the parabola (3/5,0)...the problem is that this en contains "-g" and "-f"..i.e, the co-ordinates of the centre of the circle (-g,-f)...Hence, it is not possible to obtain a non-trivial solution. I need another convinient linear in "g" and "f" to get my answer...

Please help.

SN1 and SN2 are condition specific mechanisms. SN1 is a 2 step process. and SN2 is a one step process. the reason for this is, that the cpds. that undergo SN2 rxn have relatively unstable intermediates, moreover, there is no solvation of ions, hence to avoid the reverse rxn. the process is one stepped. SN1 rxn takes place when the solvent is polar-protic (i.e. it contains H+ ions and is capable of solvating ions and that for SN2 is polar- Aprotic. Also, there is a competition in Subst. and Elim. rxn. The former is favoured at lower temp. and the latter at higher temp.- Check the question for temp. conditions if given in the question)

Secondly the attacking species could be a strong/weak nucleophile. (note that when a base attacks a substrate, SN2 or E2 mech is followed) A weak base favours SN1 as its ions can easily be solvated in comparison to the stronger ones that MAY react with the solvent.

thirdly, the substrate matters. If the carbocation formed is relatively stable, then the mechanism followed is SN1, provided appropriate rxn conditions are maintained. If the cation formed is unstable, then SN2 or the respective elimination takes place.

The stability of the cation is scaled bt the nature of the grps attached to the cationic carbon. Ex: +I grps like alkyl grps will stabilise the carbocation and -I grps will destabilize it. Similarly check for Resonance factors, Hyperconjucation etc. in the correct priority order i.e. Resonance> Hyperconjugation> Inductive effect.

Hope this has helped you. Rate if satisfactory.

i need to know the principle behind homogenising 2 or more curves. I understand that a homogenous curve gas a fixed degree in each of its terms but the concept of "homogenising" curves is not understood by me.

for ex. consider a parabola : y²= 4ax,

say a chord PQ (whose mid-pt. is [h,k]) subtends an angle of 90 degrees at the vertex. (A).The eqn. of the chord PQ is,

ky-2a(h+x)=k² - 4ah .........(1)

the combined eqn of the lines AP and AQ is obtained by "homogenising" the eqn. of PQ with the parabola.

the process used is,

since, y²=4ax,

y²=4ax [(yk-2ax)/(k²-2ah)]................(since [(yk-2ax)/(k²-2ah)]=1)

(i saw this ex. in Arihant's co-ord geom. by SK goyal)

on simplifying this, we get the desired eqns..

my question is, WHAT IS "HOMOGENISING" HOW DO WE KNOW THAT "HOMOGENISING" GIVES US THIS EQUN. HOW DO WE HOMOGENISE 2 GENERAL CURVES. WHAT IS THE INTERPRETATION OF THE RESULT OF THIS PROCESS?

please reply.

nice.

H-N=N⁺=N^{- (HN3)}

if thr proton is removed, then the N atom to which it was attached recieves a pair of electrons which are delocalized to the adjacent N atom with formal charge +1, as a consequence of which conjugate base is N

N=N^-

This will undergo solvation and hence become stabilized...

is my logic proper??? PLEASE CORRECT ME IF I'M WRONG!!!

but the same does not apply to 2nd, 3rd.... derivatives either.

think of dy/dx as rate of change of y w.r.t. x. you can perform math operations on dy/dx but not on d²y/dx² because is it the rate of rate of change which is generally non-linear in nature.

HOPE THIS HELPS YOU>>RATE IF SATISFACTORY

Linear Impulse is defined as the F_average.time.

It is intersting to know that the dimensions of impulse are the same as those of momentum.

As our forum expert rightly said, it is calculated on basis of change in momentum. Note that the Force here is the average force and not the instantaneous force. It is also important to realise that the time of collision has to be very small for impulsive forces to operate.

_{[t ]}

_{[0 ]} F_average.

t

For angular impulse, an analogous form

Impulse_ang.= Torque_{avg. about axis of rotation}.

t

the same condition of time being bery small applies here too.

For example: Consder the classic example of a disc, hinged at the top-most point.A particle strikes it at the lowermost point.

It is impotrant to note that we cannot conserve momentum because the normal force recived by the hinge is extremely large, which was a consequense of angular impulse...However we can obtain Average force with the above equations..

Hope you found this useful...Rate if satisfactory.