New answers tagged

-1

I hear this regularly. I think its a combination of the effects above. The first effect is a front of compressed air being forced ahead of each of the trains. But then, as the fronts of the trains meet and pass, the Bernoulli effect leads to lower pressure between the trains. But this isn't uniform, each carriage has its own mini air front, and the carriages ...


2

Yes, sound waves in a gas, liquid or solid can affect the light passing through it, as the motion of the atoms due to sound waves changes the atomic spacing, and this changes the index of refraction slightly. So the light would be diffracted and some amount of the light would experience a frequency shift up and a frequency shift down by the sound wave ...


6

Any physical phenomenon is potentially capable to cause some change to any other phenomenon, more or less directly. If it was not the case, the physical world could be divided into completely independent realms; there would not be the one single world we call Nature. Practically though, many if not most of the actually existing interactions between systems ...


5

I can answer half your question in that a sound can change the path of light. A change in the density of the air produces a change in the refractive index of the air and so a Schlieren photograph can make this visible. Here is a YouTube video to show a sound wave produced by clapping.


6

My high school physics teacher was saying that “this is because of interference of sound waves. During the day, there are a lot of sounds and they cancel each other due to interference. But, during the night, there are few sounds and they can reach to our ears without canceling each other”. You need a better high school physics teacher. Temperatures tend ...


-1

If we suppose that the phenomenon you describe is related with wave interference. A wave is a kind of mechanical disturbance in the medium through which it is travelling. A sound wave consists of areas of relatively high and low energy, in the form of relatively high and low pressure. To understand how sound is produced, consider a speaker. The cone or ...


3

I would tend to agree that background noise is a factor, but rather than reducing, adding to the sound you are trying to make sense of. So part of that may be how your brain is able to filter the information from the background noise. But at night the temperature is lower and according to this tutorial on sound propagation (which does cite reliable ...


22

You don't. You actually hear the high frequency notes from headphones. The bass really doesn't travel at all well, but the attack noise from the drum or bass guitar is what leaks from headphones. This is why on the tube you hear "tsss tsss tsss tsss" and very little else. From @leftaroundabout's answer on the post that valerio92 linked: Normal ...


1

Yes, the two approaches are equivalent. As you've noted, there's only one physical effect going on here, interference, and the standard beat frequency / path length interference formulas are just special cases of the same thing. There are some restrictions. Both beat frequency and path length interference only make physical sense when the sources are ...


2

The exercise seems not too difficult. With $f_0 = 456 \text{Hz}$, $c$ the speed of sound an $v$ the speed of the observer, you just have to find the beat frequency by adding the wave equations for 2 different sound waves with frequencies $f_1=f_0(1-\frac{v}{c})$ and $f_2=f_0(1+\frac{v}{c})$. So this is just an exercise in applying the Doppler effect, and you ...


0

Helmholtz effect occurs when pressure outside the box (i.e. external pressure change). There will be a whistling at certain frequency. The controlling parameters are aperture size and volume of the box. When a loudspeaker inside the box radiates, it just radiates with wave bouncing on the walls and some emits out of the aperture. Resonance occurs at ...


0

Additionally, the fundamental frequency can be obtained by: $$f_0 = v_s/\lambda_0$$ Where $v_s$ is the speed of sound, and $\lambda_0 = 4L$ where $L$ is the length of the air column in the tube. so, $$f_0 = v_s/\lambda_0 = \frac{v_s}{4L}$$ and so, $$f_n = \frac{nf_0}{4L} = \frac{n}{4L}\frac{v_s}{4L} = \frac{nv_s}{16L^2}$$


0

Water in a glass cup can be thought of as a one end closed organ pipe. The more volume of water you add, you effectively increase the length of the water column thereby reducing the length of the air column. The frequency is dependent on the length which can be understood from the following diagram. As you can see the length of the air column changes ...


0

By changing the amount of water in a tube you are changing the effective length of the tube. When you create a sound wave in the air inside the tube, the wave is reflected at the water surface and at the open end. With waves initially traveling in both directions inside the tube, they interfere with themselves and form standing waves. These standing waves ...


0

In a complex number $z=a+i b$,the real part is $a$ and abs is $\sqrt {a^2+b^2}$,. I am not an expert of acoustics but in general, when pressure is given in complex quantity and you want to measure its magnitude then take absolute. If you are comparing two complex numbers then equate real with real and imaginary with imaginary.


1

Difference between real and absolute value in general: Look at count_to_10 's answer. For acoustics and preasure measurement: Absolute pressure - pressure against perfect vacuum. Real pressure: Usually defined as the pressure against a reference-environment. Also called differential pressure. For example the pressure of the air inside a football against the ...


1

Based on some experience in music, I realized that the relation of the frequency of two pitches with equal steps say from C to D (whole step) or D to E (same whole step) is $f_2 = af_1$, where $f_2$ is the higher pitch, and $f_1$ is the lower, and $a$ depends only on the distance of steps from the lower to higher pitch. This means that the frequency ...


0

The context of the question is acoustics, but that is irrelevant. Essentially the same question could be posed in chemistry, biology, economics, etc. No physical laws or concepts can solve this problem. It is really a question about mathematics and the definition of "percentage difference." The same issue crops up regularly in mathematics homework ...


2

Keep in mind that frequency is both the wave property that is preserved in a change of medium (both wavelength and velocity change) and the physical property of sound waves that we experience as pitch. So the frequencies you hear in both cases are ones produced by the vocal cords. Nor do we expect the gas environment of the vocal cords to have a large ...


3

To a very rough approximation we can say that frequencies of speech are selected by standing waves in the speakers mouth, larynx etc. If they breath helium the speed of these standing waves increases but their wave length, being constrained by dimensions of their body, remains the same. This results in higher frequency sounds produced. (think $f=v/\lambda$...


-1

The higher temp. implies higher speed for the molecule, so it collide with the next molecule at a faster time even if they are far away from each. on the other hand, the lower temp. means lesser speed and so it can also collide with its close neighbor at a longer time. merci!


0

In a way, the 'similar crystal structure' IS responsible for ringing. This is because homogeneous solids can transmit sound over long distances, and one can excite resonances of the entire solid object, often audible. We cannot hear the ultrasonic ringing sound of a very small object, because the resonances are all above the frequency sensitivity range of ...


2

Higher frequencies are attenuated (absorbed) more strongly than lower frequencies. Here are a couple of attenuation values from this table of sound-wave attenuation (90% relative humidity): f (kHz) a (dB/km) 1 5.3 2 9 4 20 8 63 The discharge itself will generate a wide spectrum of frequencies, from infrasonic all the way ...


1

Yes. Let's consider the phase shifts to be random. Then we can consider each amplitude to be an independent random variable $A_i(t)$. Each one has a variance of $$\text{var}(A_i) = \langle A_i^2 \rangle - \langle A_i \rangle^2 = \langle A_i^2 \rangle \propto I_i$$ because the average amplitude should be zero, and intensity is proportional to amplitude ...


0

You cannot prove that because it is not generally true, or because it depends on your definition of "equal loudness". Consider the case where the sound sources are inside a flaring pipe with a standing wave (say, a modified trumpet). A standing wave is typically drawn as a standing wave of the air displacement in the pipe (illustrated), but note that the ...


1

I remember working this out in the opposite direction, hoping to get a paradox: that the amplitude of the sound produced by a choir of 100 singers is only 10 times the amplitude of the sounds produced by one of them. Each individual sound can be represented as a vector in 2-dimensional space, with length 1 (since all are equally loud) and random phase. The ...


1

The most obvious answer is you measure the total sound energy that you released and is carried away with the sound and the unit is Joule. However, this does depend on intensity, the higher the intensity, the more energy per second and the more energy/sound is released in the 10 minutes. If you want your measurement to be independent of the intensity, you ...


0

When the train moves, it displaces air in front of it and on the sides. So basically what you are hearing is the impact caused by the "air fronts" displaced by the two trains slamming against each other. You need two air fronts to hear the sound, so there must be two trains. Also, the speed must be high enough to significantly compress the air: this is why ...


-1

It is worth to mention two things. Sound is pressure wave traveling in air. When train is in motion, it brings motion to the air by the train. And when air velocity increases, the air pressure decreases based (can be seen in Bernoulli equation). When two trains move opposite to each other, the air flow is enough stronger if it is not doubled. The ...


1

I am going to steal from an answer to another question. ... a variety of sounds are heard following a lightning strike is not due to dispersion, but rather the multiple branches of the pre-strike, the main strike, and the extended distances covered by the lightning, plus, sometimes, echos. ... In the quote there is a link to a page where they ...


0

I agree with Lubos Motl that your concern really seems to be the time delay between emission and detection (which depends on distance), rather than the Doppler Effect (change in frequency, which depends on relative speed between observer and object). As Lubos Motl suggests, using light instead of sound would solve your problems. The transit time is much ...


4

The effect you noticed is a function of strength and size. When you have a large window, and a sonic boom comes along, a relatively small pressure difference can set up very large tensile forces in the surface of the glass (especially if the shock wave cannot easily "go around the back" of the glass). When there is a bending stress, all you need is a small ...


0

Or in short, it is harder to make liquid vibrate than the air because they have different qualities. If you test a subwoofer right beside you, or far from you, you can measure how long it takes for the bass to reach your ears. If you try doing it in water, let's say you have a subwoofer built for, and to submerged underwater, it will be engineered to have ...


1

Assume that the sound you are interested in is heard at 1000 Hz in air, where the sound is traveling at 343 m/s. If the device generating the sound (some type of waterproof speaker) is placed underwater, it will still generate a 1000 Hz sound, because the frequency is determined by the device that generates the sound, not by the speed of the sound in a ...


1

Electromagnetic waves are produced by oscillating charged particles but they do not need other particles to propagate. Indeed electromagnetic waves are solutions of the Maxwell equations with no sources, i.e. in the vacuum. On the other hand, mechanical waves need an elastic medium to propagate, regardless of being transverse, longitudinal or mixed waves. ...


0

Longitudinal electromagnetic waves do not exist in vacuum because the Divergence of E, and B are zero. The consequence of this is that the k-vector, propagation direction, is orthogonal to E and B.


0

Echo and Reverberation are both types reflected sound. Echo typically refers to a single reflection that's from a surface that's large compared to the wave length and that produces a specular reflection (in contrast to a diffuse reflection). It creates a visible peak in the impulse response that's concentrated around a single point in time and doesn't ...


0

Its like warm and hot. They are pretty much the same. Reverberation is the echo you get in a small enclosed space. Echo is all echoes.



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