wavelength I will call lambda, frequency I will call f If f and lambda are the same then the velocities of the waves would be the same becuase v= lambda*f You know nothing about their phase angles or the amplitude of the waves though.
The product of (frequency) x (wavelength) is always the same number ...
the wave speed. So if one of them changes, the other must change in the
opposite direction, to keep the product constant.
Speed = (frequency) x (wavelength)
Frequency = (speed)/(wavelength)
Wavelength = (speed)/(frequency)
the speed of light = wavelength x frequency
so
frequency = speed of light/ wavelength
of
frequency = 3.00 x 10^8 / wavelength
speed = frequency x wavelength
The product of (wavelength) x (frequency) is always equal to the wave's speed.
The wavelength and frequency of any wave are inversely proportional. Neither of them is related to the wave's amplitude in any way.
Amplitude, speed, and wavelength or frequency. (Wavelength and frequency are related by the wave's speed.)
the velocity of a wave is given by frequency*its wavelength
(frequency) multiplied by (wavelength) = (speed of the wave)
No. They're related by the definitions of the wave's speed, wavelength, and frequency.
(Wavelength) x (Frequency) = (the Wave's Speed).
speed = frequency x wavelength
they are related by the equation velocity=frequency*wavelength
velocity = frequency x wavelength
The product of (wavelength) x (frequency) is always equal to the wave's speed.
The wavelength and frequency of any wave are inversely proportional. Neither of them is related to the wave's amplitude in any way.
frequency = speed of wave / wavelength so if speed is constant then frequency varies inversely with wavelength
Energy of light photons is related to frequency as Energy = h(Planck's constant)* frequency Frequency = velocity of wave / wavelength So energy = h * velocity of the wave / wavelength
Amplitude, speed, and wavelength or frequency. (Wavelength and frequency are related by the wave's speed.)
Frequency of the a wave equals its velocity divided by its wavelength.