Question

The frequency of a wave is given to be $40\times {{10}^{6}}Hz$. The period of the wave will be.

Answer 1

As a first step, one could read the question well and thus note down the given values. Then you could recall the definitions for each of these quantities and thereby find the relationship between them. Then you could simply substitute the given frequency into this relation and thus find the answer.

Formula used:
Time period,
$T=\dfrac{1}{f}$

Complete step-by-step solution:
In the question, we are given the frequency of a particular wave to be $40\times {{10}^{6}}Hz$. We are supposed to find the time period for the same wave.
Before going into the answer let us first define each of these quantities and thus find the relation between them and then solve the problem by mere substitution.
Time period could be defined as the time taken for one complete cycle of vibration to pass a particular point. The SI unit of this physical quantity is seconds. Frequency could be defined as the number of waves passing a particular point per second.
One may recall that as frequency increases, the time period of the wave would decrease, that is, they have a reciprocal relationship with each other. Mathematically,
$T=\dfrac{1}{f}$
On substituting the given frequency here we get,
$T=\dfrac{1}{40\times {{10}^{6}}}$
$\therefore T=0.025\mu s$
Therefore, we found the time period of the given wave to be $0.025\mu s$.

Note: The relation for time period in terms of angular frequency is given by,
$T=\dfrac{2\pi }{\omega }$
Now, another significant quantity defining a wave is wavelength. This quantity could be defined as the distance between corresponding points of 2 consecutive waves.