@sammygerbil I had actually misinterpreted the end part of the question which asked to multiply the result by 5 and I had spent hours trying to get 365,thank you Sir.
A cylindrical tube, open at one end and closed at the other, is in acoustic unison (resonance) with an external source of sound of single frequency held at the open end of the tube, in its fundamental note. Then the displacement wave from the source gets reflected with a phase change of pi at the closed end. Why is this so? If the pressure wave gets reflected without a phase change why will the displacement wave which has a constant phase difference of pi to it get reflected?
@NehalSamee not sure to be honest, but it should normally be obvious if the lens is converging or diverging i.e. whether you should take the focal length as positive or negative.
@NehalSamee Oh wait, no, the equation does work.
If you flip the lens round then the values of $R_1$ and $R_2$ change signs, but they also change places. Remember that $R_1$ is defined as the radius of the surface the light hits first.
So in the question we have $R_1 = +20$ and $R_2=+60$
If you flip the lens round we have $R_1=-60$ and $R_2=-20$ so the equation still gives the same value for $f$.
@harambe Two Blocks on a Spring : Up to a limit the acceleration of both blocks is the same. The limit is that the maximum downward acceleration for the upper block is $g$, assuming it is not stuck to the lower block. The upward acceleration is not limited. The restoring force is not the same : when 2 blocks accelerate together, the resultant force on each is proportional its mass.
There is plenty of scope for confusion, because the restoring force is a combination of gravity force, contact force and spring force.
I thought you were being pendantic. However I guess you were trying to work out when the upper mass loses contact with the lower. You can think of the two masses as one, until the SHM downward acceleration becomes greater than $g$.
To work out the constants $A$ and $\phi$, or in the second case $A$ and $B$, you need to know the initial conditions. For example you could use the position and velocity at time $t=0$.
Though in most questions the examiner won't be so mean as to give you silly initial conditions. Typically the time zero will be taken as when $x=0$ or when $x$ is a maximum or minimum.
In this case we know $x=0$ when $t=0$ so if we're using the first form we know $\phi=0$ and we just need to find $A$.
To do this differentiate to get:
$$ v(t) = A\omega\cos\omega t $$
And putting $t=0$ gives us $v(0) = A\omega$
So if $v$ is negative at time zero that means the constant $A$ must be negative at time zero.
@Hema Reflection of Sound Waves : There is no phase change on reflection at the closed end, but there is at the open end. Pressure and displacement differ by 90 degrees, not 180. See hyperphysics.phy-astr.gsu.edu/hbase/Sound/reflec.html. ... I'm not quite sure what you mean. If still confused please post again.
Any optical diagram is reversible. If you take some arrangement of lenses with light moving from left to right then the diagram is exactly the same if we reverse the direction of the light so it moves right to left.
When you do this you swap the object and the image round.
@JohnRennie I am solving a question but still confused. The question says that one of the particles is located at right extreme, the other particle is at mean position and moving left
That surprises me. The sign convention here is really simple. One direction is positive and the other negative. Sign conventions in optics confuse the hell out of me :-)
Now put the plate in. Suppose the light ray travels a distance $d$ through the plate and a distance $x$ through the air. The the total optical length is $L = nd + x$
The point is that in the glass slab the wavelength reduces to $\lambda/n$ so in effect the light travels more than 1cm when it passes through a 1cm glass plate.
What exactly do people find so confusing about optics sign conventions? At least when it comes to distances, I find it pretty clear. Everything whose sense coincides with the sense of light propagation is positive and negative otherwise, no?
To be clear, I'm not saying I like (geometrical) optics. In fact I find it rather boring since it's mostly arithmetic and geometry.
@sammygerbil the pressure wave reflects off the closed end with zero phase difference. Since sound can be expressed either as a pressure or as a displacement wave, where the pressure maxima correspond to displacement minima and vice versa, doesn't this mean that the displacement wave also reflects off with zero phase difference? But my book says it reflects off with pi phase difference.
@NehalSamee I presume you are told the extension, so you can calculate Young's Modulus $E$. This is related to bulk modulus $K$ via $E=3K(1-2\nu)$. See conversion table at the bottom of en.wikipedia.org/wiki/Bulk_modulus.
Why is it that when a transverse wave is reflected from a 'rigid' surface, it undergoes a phase change of $\pi$ radians, whereas when a longitudinal wave is reflected from a rigid surface, it does not show any change of phase? For example, if a wave pulse in the form of a crest is sent down a str...
@IceInkberry I didn't ignore your method, I am going to read it at night. Please dont think I ignored it. I was busy with some other questions so didn't get time.
@Abcd Sorry I don't want to check your calculation step by step. When you get to the end, if your result doesn't make sense and you cannot find why, please ask me then. Haven't you discussed this problem with John Rennie a few days ago? Have you referred back to what you did then?
@Abcd suppose it goes down a distance x, the the mass has fallen a total distance of $h+x$ so the PE change is $mg(h+x)$. This must equal the PE of the spring so $\frac{1}{2}kx^2 = mg(h+x)$
@sammygerbil Okay sorry nevermind. I haven't discussed SHM with JohnRennie since almost 1 year ! (Except 2-3 sine waves problems 2 months ago). (I think you confused harambe for me)
For the upwards journey: the mass hits the top of the spring with KE of $mgh$, then it goes down, rebounds and comes back to the same point with equal and opposite velocity, so on the upward leg its KE is also $mgh$.
It moves up some distance $x$ so the spring PE is $\tfrac{1}{2}kx^2$ and the gravitational PE is $mgx$. At the topmost point the KE is zero, so we get:
@NehalSamee I am not sure what you mean. Mass of the cube can be neglected if much smaller than the load, which is used to calculate stress. Length is used to calculate stress. Stress and strain are used to calculate Young's Modulus.
Hey people, I need a double-check here. Just tell me the result you arrive at, I get a different answer from the one in my textbook, which is quite weird since my friend and I have results that match: What is the minimum angle with a vertical wall that a ladder (or a block...) can make such that it does not slip, given $\mu_1=0.4$, the friction coefficient with the floor and $\mu_2=0.1$ with the wall?
@sammygerbil Should I share my book's theory related to this part? So you can explain that? I am unable to understand some parts that the author has mentioned
@sammygerbil The difficulty I am facing is in picture 2. He says "the magnitude of magnetic component is decreasing for red"...but it appears to be increasing in the diagram.
I am unable to understand how he has applied Lenz's law to the sine loop.
@Abcd Ah I think what you need to know is that $E \times B$ lies in the direction of propagation. So you can use the Right Hand Corkscrew Rule to turn $E$ onto $B$, and screwing forward gives the direction of propagation. ...
@sammygerbil The difficulty I am facing is in picture 2. He says "the magnitude of magnetic component is decreasing for red"...but it appears to be increasing in the diagram.
What is shown in the diagram is a representation of the wave at one instant in time. This whole wave pattern moves to the right. If you look at the point where the red segment is, and imagine the wave pattern moving right, the peak of the B wave moves further right, so the B field at this point decreases.
I am tempted to say, in the same way as for a wave on water or on a string.
On a string there is a tension force between one segment of string and the next, so the motion of one segment affects the next.
I guess that in the same way there is tension or stress in the electromagnetic field, which enables the disturbance to be transmitted from one place to the next. But I don't understand this well myself.
That is the part I understand the least about. If there were an ether to transmit EM waves, then it would be a lot easier to make the analogy with mechanical waves in water or on a string. But there isn't!
Maybe pressure waves in sound are an better analogy.
And you can also appeal to momentum : light waves have momentum which keeps them moving forward.
@sammygerbil Well then I guess the answer in my textbook is wrong. Thank you very much for your time, having a confirmation from someone with more experience feels good :D
Well yes I suppose so. But we usually accept such things because of what we see in nature, eg circular waves radiating outwards from a disturbance on a pond.
Can you think of any reason why it might not be spherical?
If the wave travelled faster in one direction than another (eg vertical vs horizontal) then we would get an ellipsoid. But we assume that space is isotropic (the same in all directions). Also homogeneous (the same properties in all places). So the symmetry argument is based on such assumptions.
To have a wavefront you must have a travelling wave. This is not an expression for a travelling wave. You need something like $A\sin(\omega t-kx)$.This expresses what happens at various positions with the times at which they happen.
But we could instead choose a trough, or a zero at which the wave is rising or falling.
The shape of the wavefront depends on the nature of the source.
For example, a point source in a pond generates circular wavefronts.
But if you had a very large wall which vibrated, it would generate wavefronts of sound which are planes parallel to the wall, moving away from the wall.
Good question. The problem doesn't say what happens to the bullet. So we have to make a guess. Usually in such problems the bullet is embedded unless it is stated otherwise.
