can anybody give me the truthness of the eqn a= -(w*w)*x in SHM motion?
610689
If x is +ve then a is -ve!!!Satisfy yourself with the equation.\nThe equation means that the acceleration is proportional to the displacement but in opposite direction.\nConsider a simple pendulum.When it is at the extreme position it has maximum tendency to go back to the mean position.That tendency is the restoring force which acts opposite to the displacement and exhibits itself as acceleration.If the pendulum is displaced to half the extreme position the tendency to come back is less.Thus the acceleration is proportional to the displacement but acts in the opposite direction.\nThe equation is true in all its sense.
277032
I assume you are talking about any wave equation, not just a QM probability density function.\n\nIn any case, what do you mean that you have a "different x for every instant t"?  That is not the case.  I suspect you are thinking of travelling along with the crest or trough of a wave.  As t changes, the location of the crest and trough changes, but that is not the only location on the wave.\n\nWhat the wave equation specifies is the magnitude of the wave at *any* location x, at any time t.  So, if you are at a constant location x', the magnitude of the wave will vary periodically in time at that location. If you were a cork on the water, you would bob up and down at your location as the waves passed you. But similarly, if we were to somehow freeze time, and travel up and down our x axis, the magnitude of the wave would vary periodically in space as well.  So now, rather than a cork bobbing on the water, imagine you're on a jet ski that is going across the water so fast, compared to the speed of the waves, that it's almost like the waves aren't moving.  Yet you bob up and down as you cross the waves that lie in your path in space.\n\nHence, the wave varies both in x and t, independently.
246299
h= (C*I/A+T/k)*z*log(.7*y)
733015
a. Applying conservation of momentum\nm1 x V_i + m2 x 0 = (m1+m2) x V_f\nV_f = (m1 x V_i)/(m1+m2)\n\nb. If m1 >> m2 then m1 + m2 -> m1\nTherefore V_f = V_i\n\nc. If m1 << m2 then m1+m2 --> m2\nTherfore V_f = (m1/m2) x V_i\n\nd.m1 x V_i - m2 x V_i = (m1+m2) x V_f\n=> V_f = ((m1-m2)x(V_i))/(m1+m2)\n\ne. m2 = m1, m2 =(1/3)(m1)
708667
I think you mean\n\nS = A - RT/(1-R)\ntake A  to the left and negate\n\nA- S = RT/(1-R)\nmutiltply by (1-R) on both sides\n\n(A-S)(1-R) = RT\nor (A-S) -R(A-S) = RT\ntake R term to the right\n\nA- S = R(T+A-S)\n\nso R = (A-S)/(T+A-S)
724685
r-s = t\nadd s to both sides\nr = t+s
736379
for this just substitute for x and y in eqn and see if LHS=RHS\nLHS=4\nRHS=(-1/2)(-3)-5/2\n=3/2-5/2\n=-2/2\n=-1\nnot = RHS\nso pt. not on line
108082
Using quotient rule, chain rule, and the knowledge of the derivatives of transcendtal functions(here the exponential function), you will arrive at:\n\ny'=1/(cosh(x))^2\n\nwhere cosh(x) is the hyperbolic cosine function - \n\ncosh(x) = (1/2)*(e^x+e^(-x))\n\nThe details of the work get kind of messy, which is why I left them out, but it is really just a standard quotient rule problem. Review your quotient rule, and you should have no problem.
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
303805
I don't see how components are necessary.  \n\n"r" and "v" are just shorthand for r(t) and v(t), assuming we are looking at position (r) and velocity (v) as time varying "vectors".  And v(t) is just the time derivative of r(t) (by DEFINITION), that is, v(t) is DEFINED as r'(t).\n\nSo there isn't any actual CONTENT in expanding them into components. We are just choosing different ways to symbolize the quantities involved and there isn't anything to prove.
856976
Extreme position! \nBecause acceleration is directly propotional to the displacement from the mean position and directed opposite to it.\n\na=-(ω^2)x
808209
Using one of the equations of motion\n\n(v*v) - (u*u) = 2*a*s\n\nwhere \nv is final velocity\nu is initial velocity\na is accleration\ns is displacement\n\n\nOR \nuse \n\nv * (dv / ds) = a
150092
t=to\n_____\n√1-v2/c2
943382
If l,w and y are separate variables then\n(lw - lwy)/lw=lw/lw - lwy/lwy = 1 - y\n\nIf lw and lwy are each a vaiable then\n(lw - lwy)/lw= lw/lw - lwy/lw = 1-lwy/lw\n\nIf lw and y are variables then\n(lw - lwy)/lw = lw/lw - lwy/lw = 1 -y
823386
Hi Alan the 1st!\nWe neglect the central aperture in CD, then inertia momentum is I=m*r^2/2 for any homogeneous disk, m=0.017kg, r=0.06m. Torq T=I*w’, w’ is angular acceleration, while w=w’*t, w=22 rad/s, t=0.88s, thus T=m*r^2*(w/t)/2, - plug in the proper numbers, be a good boy!
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
698608
Have you been given the standard formulas?\nx = a cos (nt + c),  frequency = n, period = 2*pi/n, phase = c (Not sure of that -- it may be -c.  Look it up.) This gives you the answers to (d), (e) and (f) immediately.  [No, I'm not doing it for you: You can substitute values I'm sure.]\n\nx is the displacement, so just sub t = 2 in this equation.  If you need touse your calculator (you shouldn't!), make sure it's set in radians.\n\nVelocity is dx/dt.  You can do that I hope, and get\n\nv = - 18*pi*sin(3*pi*t + pi/3)  Notice I omit units.  I prefer to use equations relating to numbers only (though I know units are sometimes useful especially in a Physics context) and then state the correct units in presenting the answer -- which you are going to do!\nSub t = 2 to get v in m/s\n\nAcceleration is dv/dt, so differentiate and then sub t = 2\n\nAll the best