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The frequency of the first harmonic oscillation is the higher the higher the tension in the string. As temperature increases, the length of the string slightly increases. The change of linear mass density is thus negligible, but the corresponding change of tension in the string is not. The tension decreases and thus the speed of waves and frequency of ...


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To find the minimum velocity at the bottom-most point, we find the minimum velocity at the uppermost point. This minimum velocity is the one such that the centripetal force is equal to the force of gravity. Any lower and gravity will pull the object down and out of the loop, any higher and a faster velocity would be required to generate it (thus, not a ...


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A tension less than zero isn't really physical; this is the point where the string stops being taught and the object doesn't make a complete circle. Therefore, when finding the minimum velocity for an object to make it around a loop, we solve for when the tension at the top is zero as it is the minimum possible tension for the object to keep going in a ...


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here is an animated version of image for more in formation refer to http://www.physicsclassroom.com/class/waves/Lesson-3/Boundary-Behavior and a good tool for playing with https://phet.colorado.edu/sims/html/wave-on-a-string/latest/wave-on-a-string_en.html


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My intuition is that the frequency should stay the same because the waves in the light rope are caused by the waves in the heavy rope. The point where the ropes attach will oscillate with a common frequency. So, for $(b)$, the frequency would be the same. For $(c)$, use the equation $v= f\lambda$. You already correctly determined that the velocity ...


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a) For the block: $Ma = T - F$, using Newton's second law. For the sphere: $ma = mg - T$, using Newton's second law. Combining these we can get the acceleration of the system: $$ a = \frac {mg - F}{M+m}$$ Now, using $v^2 = u^2 + 2as$ in the form $v^2 = 2ah$, as $u = 0 ms^{-1}$ we get: $$ v^2 = \frac {2(mg - F)h}{M+m}$$ as required. b) Considering the ...


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When you push a rope, it has that bulk property to easily change its shape and bend without producing much of a reaction force. But if you pack the rope in a very narrow hollow cylinder (an impossible ideal case indeed) where the rope won't have the freedom to bend, and then push it, it will get compressed first filling up gaps in the fiber binding but after ...


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Your situation is a little confusing since the velocity of the wave packet is not related to the momentum. If you want to speak about the uncertainty principle for classical waves, the 'momentum' is proportional to the inverse wavelength $\lambda^{-1}$. In this situation the speed of the wave does not depend on the wavelength, only things like the density of ...


4

I think you have misunderstood the meaning of the equation $$\Delta X \Delta P \geq \hbar / 2 \, .$$ This is not surprising given that the notation used here is really, really misleading. It should be written like this $$\sigma_X \sigma_P \geq \hbar / 2 \, .$$ To understand this we have to explain what $\sigma_X$ and $\sigma_P$ mean. Suppose you have a ...


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Problems that depict situations where the tensions are same on ropes on both sides of the pulley are ideal situations.It is stated so in order to minimize any complexities that may arise if the pulley was to rotate.Now, if the tensions were not equal on both sides, the pulley would experience a net non-zero torque and hence a net angular acceleration and ...


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Because the pulley possesses mass, you need to apply a non-zero net torque to it to increase its angular acceleration (assuming that is the goal here). If the tensions were the same on both sides of the contact point between the string and the pulley, there would be no angular acceleration.



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