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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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If the interaction with the wall is linear, the dynamics of the surface can only modify the existing harmonics of the incident sound wave--it can not create new ones. Re. 'fundamental frequency' https://en.wikipedia.org/wiki/Fundamental_frequency The fundamental frequency, often referred to simply as the fundamental, is defined as the lowest frequency ...


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No. I suppose it might be possible to design an exotic device which absorbed energy from the incident sound wave at one frequency and re-emitted it at a different frequency (by analogy with light phosphors, etc) but that is not the conventional meaning of "reverberation" in acoustics.


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If you are referring to the frequency of the light in the pulse there is no relation between it and the times of the pulses. But if you are referring to the frequency of the pulses, in pulses per second, you get a $\frac{1}{time}$ relationship. The period of the pulses is $ \frac{1}{frequency}$. If you measure 5 pulses in a second the frequency is 5 ...


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For a nucleus with medium mass, the first excited vibrational state is typically at an excitation energy of about 1 MeV. We can relate this to a frequency via $E=\hbar \omega$, which gives $\omega\sim10^{21}$ Hz. This is what is known as an isoscalar vibration, in which neutrons and protons move together. Vibrating molecules is not the same thing - which ...


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The frequency is $f = \omega/(2 \pi)$ and specifies the oscillation in time. The wavenumber is $\beta_z = (2\pi)/\lambda_z$ and specifies the oscillation in space. Thus the electric wave is given by $E \cos(\beta_z z + \omega t)$. The dispersion relation describes the relation between $\omega$ and $\beta$. In vacuum, this becomes $\beta \omega = c$ or $\...


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First, let me just point out that the relationship between the angular frequency $\omega$ and the propagation constant $\beta_z$ is not in general as simple as indicated by the second expression. For instance, in an optical fibre, which is a widely used waveguide, the relationship is quite complicated and generally need numerical methods to determine. Now, ...


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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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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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