# Tag Info

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Force gets generated by the interaction of the current flow through the coil and the magnetic flux of the motor. Hence you design to have current and flux to be perpendicular to each other and the motor is designed to concentrate all the magnetic flux in the gap. The current in the coil does indeed create a magnetic field as well, but that is an undesirable ...

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It is a very fundamental good question and it is not limited by acoustical waves but is also valid for all harmonic wave propagations bounded by a reflecting surface. Here it is the room, ie, standing waves are observed also for electromagnetic waves very similarly. I find it helpful to clarify some points first, especially for other readers following this ...

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Like very well the rest of the contributors commented "wave-beams" (apologies for the slight abuse of the term) are not uncommon at all. Medical imaging is just one field where they are used. Sonars is another possible application (both transmission, and reception). In general, in acoustics (whether it is ultrasound, underwater, or "conventional acoustics") ...

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The speed of sound is found (both mathematically and experimentally) to be: $$v = \sqrt{\frac{P}{\mu}}$$ . Let's understand this formula a little, it depends on pressure directly (although to $1/2$ power) means if we increase the pressure the speed will be increased because more pressure means that molecules are hitting the walls of container strongly and ...

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Beamed sound waves are indeed routinely created by the high-frequency horns used in both high fidelity speaker systems and high-powered (stadium-sized) public address systems. Their design allows either broad or narrow angular dispersion, and widely different horizontal-versus-vertical dispersion angles. Their design is described in most upper-division-level ...

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The beam width is proportional to the wavelength $\lambda$ divided by the aperture width $L$. Audible sound frequencies are are in the KHz range with wavelengths between approximately 17 m and 17 mm. Whereas visible light wave lengths are in the micrometer range. So sound apertures would have to be vastly larger than light apertures to achieve similar beam ...

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Wave beams require to have a transversal section of lenght of the same the order of magnitude than the wavelength. Whereas for light, we can get very tiny and focused beams (of $\mu m$ order), for sound the wavenlength (of centimeter or meter order), you cannot get beams s focused. Hence the utility of such beams to either transmit information, or focus ...

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Try to figure out the change in the kinetic energy of the ball after the bounce. Here I am assuming that the elastic plate remanis motionless. Now this change in kinteic energy is the energy that has been either dissipated as heat, or stored in the deformation of the ball(or plate) or(finally) transferred into the energy of a sound wave. Now I am assuming ...

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As you mentioned, the pressure difference between the outside volume and the vacuum sealed volume creates a gas flow into the vacuum sealed volume when the two volumes are joined together (i.e. there is no resistance to the flow or, lets say, air. The rapid movement of air at the boundaries of the two volumes results in a turbulent airflow, which results in ...

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They are not really the same thing, attenuation is the loss in sound energy could be due to absorption, reflection, scattering...etc. Absorption is converting sound energy into heat energy when the sound wave hits the absorptive object. So absorption is part of the attenuation but the attenuation is not necessarily absorption. For example, noise barriers ...

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The shape effect on fundamental pitch is slight, but the effect on the overtone series is significant. Most wind instruments have cylindrical shapes for this reason, and also because square cross-section pipes with bends in them are more difficult to fabricate. This is a topic about which a lot has been written in the field of musical instrument acoustics/...

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I believe the question can be rephrased (more clearly) as follows: "when we collide two glass balls (or two steel balls), we hear multiple collisions. The intervals between collisions change from long to short. Why?" Take a look of this video https://www.youtube.com/watch?v=k1id4a4EU4M (Pay attention to time 1:01) when he casually collided the balls you ...

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A vibrating object like a wine glass or a bell is a very complicated system. It is possible for a wine glass, for example, to exhibit multiple resonance modes in which the modes are weakly coupled i.e., energy in one mode is slowly shared with other modes. This means the quality of the perceived sound will shift perceptibly on a timescale of ~seconds. Note ...

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Assuming that you are driving straight into the building, the observer in the building hears the sound coming from your moving car with frequency shifted due to the Doppler effect. The frequency is given by: $$f_2=\frac{c_s}{c_s-v} f_1$$ Here, $c_s$ is the speed of sound in air and $v$ is the speed of your car. You can find the derivation of the frequency ...

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$v$ applies to the speed of sound in the equation $f=\frac{v}{\lambda}$. Assuming air to be an ideal gas we can use the following equation to calculate the speed of sound in air: \begin{equation} v=331.3\sqrt{1+\frac{T}{273.15}} \end{equation} where $T$ is the air temperature in degrees Celsius. The wavelength should be twice the length of the organ pipe, ...

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Well, Qqwy, has provided a nice answer to you, so I am just gonna try to add up to that. As has already been mentioned, the actual realization of an active noise cancellation device has many implications. Latency being just one of them (more of them include imperfections in modelling, transfer function of the components, incorrect equalization and a whole ...

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I believe that Erlend gave a quite instructive answer, so I will try to stand only on one point of your question. You kinda mix loudness, amplitude and frequency in a more objective way than a subjective one (as would the inclusion of loudness in the conversation would demand), and you seem to be interested to know if the frequency of a wave would affect (...

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Well, if you want to be mathematically correct (in an objective manner), the noise will be doubled. As niels nielsen mentioned doubling the power will yield a 3 dB increase. Now, whether this results in a doubling of loudness as a perception metric, then the answer is most probably no. According to this book and further references therein, as well as the ...

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I miss in the answers some model about what physically happens when the wave enters a new medium. There may be of course many other models, but for this particular purpose it may be illustrative to think of two theoretical possibilities: the wave is like a train with rigid wagons (the length of a wagon corresponding to wavelength) or like a line of soldiers (...

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Here are some things for you to think about. First, to reduce the resonant frequency of a metal tuning fork requires the removal of enough metal from the crotch of the fork so as to significantly increase the effective length of the tines. It is highly doubtful that you did that with just a couple of passes of a metal file. Second, to get a reliable ...

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Hold it against a surface (sounding board). Or put it on a quarter-wave resonator box. Or hold it against your teeth or on your skull behind the ear.

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Suppose you have a rope with variable mass per unit length $\rho$ along its length. We are going to perform some kind of local stretching and/or compression along the rope, and we want to quantify how the changes in linear density are related to the stretching we impose. To quantify the kinematics of the deformation, we are going to label the various cross ...

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Sound is just pressure waves propagating back and forth, so kinetic energy, when it gets absorbed by an object it just jiggles its constituent particles, so basically it stays kinetic energy; macroscopically you could say it turns into thermal energy since the thermal energy of an object is the total kinetic energy of its constituent atoms/molecules.

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Your ears detect loudness on a logarithmic scale, where doubling the strength of the noise source results in a slight but noticeable increase in the perceived loudness, which is +3dB when measured on a sound level meter. So, if one washing machine generates (for example) a 75dB noise level, adding a second washing machine will measure (75 + 3) or 78dB on a ...

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According to this article, the loudness discrimination threshold for human hearing is around 0.6 db. I've seen other articles that place it at around 1.6 db. It seems to be dependent on the frequency spectrum of the sound.

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Yes there are some nice partial explanations above of how we get reflection back at open end. Also very important is to realize this only happens when the wavelength is similar or longer than the width of the opening ... otherwise with a small wavelength the sound will propagate in the main as straight rays out of the pipe like bullets out of a gun with ...

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Your method is completely fine and it is the one that should be used. Let me write it down once more $$v = \sqrt\frac{{T}}{{\mu}}$$ $$300 = \sqrt\frac{500}{\mu}$$ $$\mu = \frac{500}{90000}$$ $$\mu = \frac{5}{900}$$ No we want $v=312$ so let's just put it in the equation which you have stated $$312 = \sqrt\frac{900T}{5}$$  \frac{312\times 312\times 5}{...

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Your last equation is for differentials, it is only exact when the differences ($dT,dv$) are infinitesimally small. For any differences that are finite the answer will be close but slightly off. What you did is essentially a Taylor expansion. If you write $T$ as function of $v$ (assuming $\mu$ is constant) you can use a value at known $v=v_0$ to get an ...

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I don't want to disappoint you, but the frequency of a plucked string is almost independent of Young's modulus, so if you are trying to measure E this is a poor way to do it! However that wasn't the question you asked. The problem with trying to use a FFT here is that the FFT works best when the amplitude of the signal is constant, and that is clearly not ...

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Usually, the stair well in a house has fairly hard reflective wall surfaces and nothing much that would absorb sound energy. Therefore, there would be frequencies at which a strong standing wave pattern would be produced in the space. If you were at a position where the amplitude of the standing wave was zero (called a "node") then you would experience more ...

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This sounds like destructive interference interacting with your attempt at driving a resonance. Normally when whistling or singing in the shower or a stairwell, certain pitches get nicely amplified by resonance. These room modes are standing waves where the pitch fits into an integer number of wavelengths between reflecting walls. The sound source ...

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It could be possible that the produced sound echoes in the room and the echo makes destructive interference with the original sound. It would affect different harmonics differently however, so I recommend also trying to use a the same note an octave higher or lower to exclude this possibility.

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