Wiki User
β 12y agoPeriod = 1/frequency = 1/250 = 0.004 = 4 milliseconds
Wiki User
β 12y agoPeriod = reciprocal of ('1' divided by) frequency = 1/500 = 0.002 seconds = 2 milliseconds
Period = 1 / frequency
0.48 KHz = 480 HzPeriod = 1/frequency = 1/480 = 0.0020833 second (rounded) = 21/12 milliseconds
Wave frequency f, and period of wave T are inverses, related by fT=1.
frequency = (wave speed)/(wavelength) frequency = 1/(period)
The period of a wave is the inverse of its frequency, so for a wave with a frequency of 0.50 kHz, the period is 1 / 0.50 kHz = 2 milliseconds.
Period = reciprocal of ('1' divided by) frequency = 1/500 = 0.002 seconds = 2 milliseconds
Period = 1/Frequency = 0.00175 seconds (approx)Period = 1/Frequency = 0.00175 seconds (approx)Period = 1/Frequency = 0.00175 seconds (approx)Period = 1/Frequency = 0.00175 seconds (approx)
Period = 1 / frequency
If the period of a wave increases, the frequency of the wave will decrease. This is because frequency and period are inversely proportional, meaning that as one increases, the other decreases.
True. The period of a wave is inversely proportional to its frequency. That means as the frequency of a wave increases, the period of the wave decreases proportionally.
When the period of a wave decreases, the frequency of the wave increases. This is because frequency and period are inversely related - as one increases, the other decreases. So, a shorter period corresponds to a higher frequency.
0.48 KHz = 480 HzPeriod = 1/frequency = 1/480 = 0.0020833 second (rounded) = 21/12 milliseconds
Period = 1 / frequency
The frequency of a wave is the reciprocal of its period, so if the period is 6 seconds, then the frequency is 1/6 Hz.
0.95 kHz = 950 Hz So period = 1/950 seconds = 0.001053 seconds (approx)
Yes, as the frequency of a set of waves increases, the period of each wave decreases. This is because frequency and period are inversely related - frequency is the number of wave cycles occurring in a unit of time, while period is the time it takes for one wave cycle to complete.