As Chi Meson stated "difference in tube length between any two successive harmonics is half a wavelength", using this u can easily find the wavelength and then u can find velocity. U will find that the lowest frequency for the tube will be in the second harmonic.
Here is the correct solution I just figured out.
fR=I(alpha) ...(f=frictional force)
maR=(2/5)mR^2*(a/R) ( where a is the translational accln.)
m(Vx-Vo)=(2/5)mR^2*((Vx-Vo)/R)
from here u will get Vx=(5/7)Vo... =p
can u give me a diagram of the arrangement because in I assumed electric field to be in the left directrion and positive charge in the left and the negative charge in the right. Is that right?
ok distance traveled from the time the ball has been thrown and it starts to roll= (Vo^2-Vx^2)/2a. Vx being the final velocity, a being the accln due to friction.
Now I apply the conservation of energy theorem
0.5mVo^2=0.5mVx^2+0.5I(Vx/R)^2+ma(Vo^2-Vx^2)/2a
after solving this equation I get...
Homework Statement
A battery has an emf=6V. The battery is connected in series with an ammeter and a voltmeter. If a certain resistor is connected in parallel with the voltmeter reading decreases by a factor of 3, and the ammeter reading increases by a factor of 3. What is the INITIAL READING...
from what I know we use Asin(wt) when u start counting ur time from the equilibrium position while Acos(wt) for time starting from the amplitude of the SHM. U don't need to worry about phase constant here since the system is starting from its mean(equilibrium position).
first of all, I think it should be an inelastic collision. If it is so,
Then as u have done above u can find the v of the system. This will be vmax. Hence equate the total kinetic energy of the system with the with the total potential energy 0.5 *k *x^2. The x u will get now is the ampitude. W u...
it is not that the "same spring will have different spring constant", but different length of the same spring will have different spring constant. Suppose u have one string of 1 m and k=1. so if u break it like 2/3 from say right. Then the spring constant of the right spring will be 3k/2 will...
no actually ur working is wrong. considering centre of mass as a reference frame, the length from the cm to m1 will be m2*x/(m1+m2) and to m2 will be m1*x/(m1+m2). now spring constant k1 for m1 will be (m1+m2)/m2*k and k2 will be (m1+m2)/m1*k. so the (omega)^2 for both will be ((m1+m2)/m1m2)...
ur answer seems to be overly complicated to me. Here is my solution:
-GMm/r=-GMm/R + 1/2 m v^2 . r being the distance between the particle and centre of the ring.
from here u will get v = sqrt. 2GM(1/R-1/r) which can be reduced to sqrt.2GM/R since R<< r. Then it easily follows the answer...
Homework Statement
I need to design a windmill in a competition that can produce the maximum power given every participant has the same gears and dynamo. so basically we have to concentrate on the blade's design. What is the best design and its orientation on the rotor. Also what type of...
if the path is a rectangular hyperbola and the particles are at infinity then how is it that the perpendicular distance between them is L. It should be 0. The equation of a rectangular hyperbola is xy=c^2. So if x→∞ y→0. So the perpendicular distance is 0.
the voltage will be the same and the electric field too provided the battery is attached to the circuit. But charge on the plate won't be uniform. the charge on the part of the plate near the k1 will be different from k2. Am I correct??
no I m not talking about capacitors in series I m asking about what if they are parallel?? I mean both of the dielectric are placed side by side nd both of their edges touch the two plates
Are the two particles parallel to each other moving in opposite directions? and I don't understand what is meant by the distance between the paths. Is the perpendicular distance between the two parallel lines??
so suppose there is two dielectric slabs with k1 and k2 placed in parallel in a capacitor. Then can we say that the resultant electric field between the two parallel capacitors now is different.
I think u should rethink on the current and conventional current. U can assume the electron moving from the left to right as the a conventional current from right to left. Now there are two wires with conventional current passing in the opposite direction. So they will repel . So the electron...
Homework Statement
This is not a problem from any book or any kind of source. I was just thinking on this. Suppose u have a capacitor attached to a battery. And then u insert the dielectric slab. Then do the electric field between the plates changes?
Homework Equations
The Attempt...
no the author is correct. For a negatively charged particle the direction of magnetic force is opposite to the positively charged particle. It will be down only when the particle is negative. F= qbv. here the sign of q also matters...
you have to do a bit of geometry here first. see the attachment. U can find r by the formula u stated. Then use sin(60-r)/sin(x)=1/n to find the angle x at which they emerge. You have to do separately for the red and violet but using the same process
u r lacking some quantities. The question should also include where is the 2nd transmitting antenna. Let it be d
Then if u find that d=m(lambda) where m=0,1,2... then its a constructive interference.
Moment on the rod due to the spring is 2kl^2*theta. Equate it with I(alpha) then u will gte alpha and hence omega. From here u can proceed easily to find the period
Homework Statement
A boat is rowed with constant speed u starting from a point A on
the bank of a river of width d , which flows with a constant speed
nu . The boatman always points the boat at a point O on the other
side of the bank opposite to A. Find the equation of the path
r = f...
Agree with Doc al. It is a straight forward question. Moreover I got a very nice answer. It is (5/9)L. ( I don't really trust the coaching institutes u r talking abt, I have sometimes find mistake in their solutions to IIT JEE question paper.)
no not really. once the protons velocity has changed direction the proton will move away and thus distance between them will start to increase. We can consider it as two carts of different masses coming towards each other with a spring attached to them.
If not so, it is really not possible to...
you just find accln. in terms of dm/dt. and a= F/m. since m is changing u have to express it in terms of M, m and dm/dt. treat dm/dt as a constant (as given) and then integrate both sides. the answer will be in natural log form.btw ur answer is bit wronf. instead of dm/dt it should be dt/dm.