Waves at low frequencies have a larger wavelength. Typically, low noises are roughly at ~75 Hz, the speed of sound is about 340 m/s, so that gives a wavelength of [itex]\lambda = v/f=340/75 \approx 4.5 m[/itex]. Since this is much larger than the typical dimensions of the obstructions the waves meet, it will 'bend around it'.
Higher notes, at a frequency of about 750 Hz, which is 10 times higher, have a wavelength taht is 10 times shorter ~45 cm. Waves at these frequencies are more easiliy blocked.
"bend around it", please explain, as I am inside a closed house so sound must travel through the walls.
Diffraction will only occur if the wavelength is large compared to the size of the obstacles.
Even though your house is closed, the sound will be transmitted through the windows much easier than through the walls. This too might be frequency dependant due to the resonance frequencies of the windows and walls, but I`m not too sure about that.
You can tell if the walls and windows are resonant with the frequency from your neighbours bass because they will rattle and shake quite noticably. In fact the rattling and shaking at my house is more annoying than the actual music.
I too have loud neigbours .
Is that somewhat similar to the resonance when you place two tuning forks beside each other and make one vibrate
no - that's either forced or the 2 tuning forks have the same natural frequency. But think about it, those idiots, that spend thousands on sound systems, aren't even benefitting from it - what kind of rice burner has ANY 4.5 meter span, let alone one, from the front of the sub-woofer. The car is too small to even set up a wave in.
Automobile cab tuning frequencies are typical based on ¼ wavelengths not the full wavelength.
I have a bunch of Menonite (sp?) neighbors. Loudest noise is the occasional car starting up around here :D Except when some low rider or whatever jerk comes up the main street a few blocks away with their stupid base.
The correct answer is that wave amplitudes are absorbed exponentially with distance
according to the wavenumber. If the wall is only 0.01 wavelengths thick, the sound
will get through much better than if it is .1 wavelengths thick.
I agree with Antiphon, and I guess you can also think in terms of the cavity (room) resonances. In a cubic room of 5 meters you have [itex] \lambda = 68 m [\itex] which is pretty close to the wave length you are mentioning. Moreover this resonances tell us that the transmissivity of you walls get higher at these frequencies. You have also to consider how much energy is concentrated in each frequency of the music source, i.e. its spectrum.