@sammygerbil aaah, yes, the question specifically says friction between all contact surfaces. Damn, I didn't read the question properly - rookie error!
@JohnRennie A biconvex thick lens is constructed with glass $\mu = 1.5$. Each of the surface has a radius of $\pu{10 cm}$ and the thickness of the middle is $\pu{5 cm}$. Locate the image of an object located far away from the lens.
@JohnRennie We just have to consider refraction from both surfaces and not use stuff like lensmaker's formula or lens formula which is for thin lenses.
If the direction of those light rays is the same as the direction of the light rays that would come from an object at the position of the virtual object the lens can't tell the difference.
The equation is telling you what happens at the lens, so it works fine for virtual objects and images because the lens doesn't care where those objects and images are. It only knows what angle the light rays pass through it.
The virtual object may be to the right of the lends, but the (real) light ray is travelling left to right and the (real) light ray defines the direction.
@Abcd correct. The virtual object is not emitting a light ray that travels right to left.
Always use the real light ray to define the direction.
If the direction of those light rays is the same as the direction of the light rays that would come from an object at the position of the virtual object the lens can't tell the difference.
@RaviPrakash there are some good videos on YT by professors.I usually prefer Mr Michwl Van Biezen videos if absolutely stuck... Maybe you can try if you want
@JohnRennie One end of a cylindrical glass rod of radius 1 cm is rounded in the shape of a hemisphere. The rod is immersed in water and an object is placed in the water along the axis of the rod at a distance of 8 cm from the rounded edge. Locate the image of the object
@JohnRennie I just can't understand the question. Just need help with the diagram.
But now, in past few years you might have came across websites MIT OpenCourseWare™, Harvard.edu, etc. so which website like it do you recommend? Cambridge? :-)
Hmm, that means $u=0$ which is going to blow up the equation ...
I'd guess we consider a small region at the point of contact. Over this region the glass surface is approximately flat so no refraction occurs and it's just a bird's eye view. So the words appear above the paper.
Suppose the blocks are glued together, then the equation of motion is the usual $A\sin\omega t$, and you can differentiate this to get the acceleration.
Or more precisely the deceleration since you're interested in what happens at the upper part of the motion.
So as long as the deceleration $A\omega^2\sin\omega t$ is less than $g\sin\theta$ the two blocks will move together.
If $A\omega^2\sin\omega t > g\sin\theta$ then the deceleration of $m_1$ will be greater than the dceleration of $m_2$ and the two blocks will separate.
@harambe yes, the block separates as soon as the deceleration > $g\sin\theta$. This will happen at some point during the upwards travel, not necessarily at the maximum height.
To help with the understanding imagine the spring is vertical with the bottom end on the floor and the two blocks on top. Are you clear what I mean or should I draw a diagram?
This shows the situation at equilbrium i.e. everything is stationary. The weight of the blocks has compressed the spring until the spring force $F_s$ equals the weight of the blocks so $F_s = (m_1 + m_2)g$
So the spring exerts a force $F_s = (m_1 + m_2)g$ on the bottom of block 1. Then block 1 exerts a force on block 2. This force is equal to $F_s - m_1g = m_2g$.
So block 2 has an upward force $m_2g$ from the spring and a downward force $m_2g$ due to gravity and the net force on block 2 is zero. So far so good. Everything is in equilibrium.
OK, now suppose the masses are oscillating so at the instant shown they are moving upwards. Then some time later the two blocks have moved up a distance $x$ (diagram 2 incoming):
The force exerted by the spring has been decreased by $kx$ because the spring has been stretched relative to its equilibrium length. So the force exertd by the spring on the bottom of $m_1$ is now $F = (m_1+m_2)g - kx$. OK so far?
So you're calculating the (downwards) acceleration in your SHM system and finding where that acceleration is larger than the gravitational acceleration.
@Abcd let's just finish off harambe's question ...
In the thin lens approximation we consider the lens to be (effectively) zero thickness so all the refraction can be treated as if it happens at the mid plane of the lens.
Two cars A and B are moving towards each other with same speed 25 m/s. Wind is blowing with speed 20 m/s in the direction of motion of car A. The car A hears horn by B as well as its ownhorn reflected from B. Both horns hav natural frequency f = 550 Hz.
How is diference in wavelengths received 365? (velocity of sound in air = 320 m/s).I did this by puting frequency perceved by A due to B's horn f1 as 325f/275, frequency perceved by B due to A's horn f' as 365f/275 and frequency perceived by A due to reflected f2 as f' (325/275) or f(365*325/315*275).
To find difference in wavelength I took as (320+20-25-25)(1/f1 - 1/f2), but that doesnt give the right answer. What should I be doing? I have tried this problem in several different ways but can't get the answer.
