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# Tag Info

## Hot answers tagged waves

2

We write $kr$ to show how the wave changes through space. For example you can fix $t$=constant, so the part $e^{i\omega t}$=constant, so you can see changes through space just shifting the "$x$" - space component. "$kx-\omega t$" express the whole phase, so you take a real part if this exponent, and after multiplying the amplitude you get the value of ...

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Short Answer Your intuition is right - masses oscillate faster for a low wavelength wave because of greater distortion; it's just that in this case the wave also repeats faster in space...so phase velocity remains the same. If you like, mathematically, low $\lambda$ means low time period $\tau$ of oscillations of a particular point on the string, so that ...

1

The exact formula for the radius of curvature is $$\frac{1}{R} = \frac{y''}{(1+y'^2)^{3/2}}. \tag{1}$$ Exactly at the minimum or maximum you have $y'=0$. Near the minimum/maximum you still have a very small $y'$, i.e. $|y'| \ll 1$. Therefore you can use the approximation $1+y'^2 \approx 1$. Then equation (1) from above can be simplified to  \frac{1}{R} \...

1

There will be a standing wave whenever you terminate a transmission line with a mismatched load. The line is resonant when the standing wave has a minimum or maximum of amplitude at the input (or feed point) of the line. If the termination is short, open, or has purely real impedance, then this occurs when the line length is $L = n\frac{\lambda}{4}$ for ...

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This could be thought of in terms of the impulse response of the wall. If the impulse response was non oscillatory (say, a decaying exponential) then the sound on the other side would be a low pass filtered version of the impinging wave--whether it is an impulse or something else. Otherwise the wave on the other side could include the resonant ...

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The extraneous springs cause faster vertical oscillations on top of what is already caused by the non-dispersive medium (the rope), thereby propagating the waves faster and making the dispersion relation non-linear. For large $\lambda$, the spring energy dominates the oscillation because the string is stretched in a rather smooth way, so less contribution ...

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