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The strength of the laser determines wether or not is beam is visible to the naked eye


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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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So in our results, i found that the frequency recorded in a quiet area was higher than a loud area, varying by around 5Hz. Can this be attributed to the interfering frequencies?


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In addition to the two fundamental frequencies you will also get the beat frequencies as they interfere in the microphone


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This peak is typical for the resonance curve of any (weakly damped) harmonic oscillator. It naturally comes as a solution of its differential equation for different frequencies. Driving the oscillator at a frequency slightly below its resonance frequency leads to big amplitude of oscillations. Therefore, even a weak electric field causes a strong electric ...


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In many cases people do seem to say "Fourier frequency" when they mean "frequency". However, when dealing with data defined only on discrete time points the phrase "Fourier frequency" has important meaning. Consider a sequence of $N$ values $\{ x_n \}$ where $n \in \{1, 2, \ldots N \}$. This situation comes up all the time if we have a physical signal ...


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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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To help you with section a: Your approach in writing the differential equation for the two positions (http://oi62.tinypic.com/dfx6yq.jpg) is correct. Since we know gravity won't affect the resonant frequency, I'll ignore gravity when writing the equations. $$\ddot{y_1} = -(k/m_1)(y_1-y_2)$$ $$\ddot{y_2} = -(k/m_2)(y_2-y_1) $$ Notice that the spring force ...


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I'll start with the second of your questions. Yes, light from very distant galaxies gets redshifted to such long wavelengths that there practically isn't any light to see. The lower limit on frequency is zero. Obviously. Technically one could say there is no signal at $0\,Hz$, but that still put a lower boundary on the frequency. Objects on the edge of our ...


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Your understanding of resonance seems about right on a qualitative level. If one were to ignore losses like friction, drag, or the like, "driving" a system at its resonance frequency would indeed result in feeding it more and more energy which is stored in the form of a large amplitude of the oscillation. For a completely lossless system, the amplitude would ...


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It depends what you mean by "function optimally". If you want it to be loud, it must transfer energy to the air and thus it's vibrational energy must decay rapidly. Imagine that the tines are large thin vanes (close together) so they transfer most of their energy to the air in a few vibrations. You will not hear or measure a precise frequency. If "function ...


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An interesting question. But more information is needed. I'm assuming this is a heater or air conditioning vent. Is there hot or cold air coming out of the vent? In either case, is there a significant temperature change as you move away from the vent (significantly higher or lower?) Although the change in temperature will affect the speed of sound as it ...


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I'm going to interpret your question a bit liberally - you ask for the case where we ignore the Sun; I'm going to go a little bit further and ignore the entire galaxy (and in fact other nearby galaxies) and talk about the cosmic background radiation. The cosmic microwave background gets a lot of attention, but in fact there are cosmic backgrounds at a very ...


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EDIT: Now that the question has been reformulated, this isn't the actual answer (it gives the value of the average detected wavelength) Two assumptions: We're looking for some average values in the cosmos, not the specific value we get here on Earth (since the sun messes the whole calculation up) We're looking for the average wavelength, not the most ...


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Run this to determine which frequency range(s) you can hear http://onlinetonegenerator.com/hearingtest.html If you can hear 14kHz-15kHz then the problem is with the radio or transmitter otherwise it is your hearing I imagine that loss of hearing at an intermediate frequency would be rare especially if it is in both ears. My bet would be with poor speaker ...


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First, check your algebra. You actually end up with units $\sqrt{\frac{m\cdot Pa}{kg}}$. Now, one Pascal is one Netwon per meter squared, $Pa=\frac{N}{m^2}$. One Newton is one kilogram-meter per second squared, $N=\frac{kg\cdot m}{s^2}$. Thus one Pascal is $Pa=\frac{kg}{m \cdot s^2}$, and if we plug that into $\sqrt{\frac{m \cdot Pa}{kg}}$ we get units of ...


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For example, when you are looking at the photon emitted from an electron going from energy level n=2 to n=1, the photon has a discrete energy and all photons that do this transition will have the same energy. However, if the photon passes from different energy levels (say n=3 to n=1), it will have a different energy. Both photons have discrete energies; ...



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