Answer
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Hint: Frequency of oscillation is the number of oscillations in one second
The relation between frequency and time period is given as
$F = \dfrac{1}{T}$
Where, $F$ is the frequency and $T$ is the time period.
The time period is given as
$T = \dfrac{{2\pi }}{\omega }$
Where, \[\omega \] is the angular frequency
Complete step by step answer:
An oscillation is a back and forth motion. In the case of a simple pendulum one oscillation is one complete cycle of swinging and returning back to its original position. The time taken to complete one oscillation is taken as the time period of the pendulum.
We know that frequency of oscillation is the number of oscillations in one second. Frequency is expressed in the unit Hertz. Its symbol is $Hz$. This can be calculated by taking the reciprocal of time period.
The relation between frequency and time period is given as
$F = \dfrac{1}{T}$ (1)
Where, $F$ is the frequency and $T$ is the time period.
We know that time period is inversely related to the angular frequency.
$T = \dfrac{{2\pi }}{\omega }$ (2)
Where, $\omega $ is the angular frequency.
Here, $2\pi $ is in radians. And $\omega $ is measured in radians per second.
Now let us substitute the value of time period as expressed in equation (2) in equation for frequency.
Then we get,
$F = \dfrac{1}{{\dfrac{{2\pi }}{\omega }}}$
$\therefore F = \dfrac{\omega }{{2\pi }}$
This is the relation connecting frequency of oscillation and angular frequency.
So, the correct answer is option C.
Note: From this relation we can see that the angular frequency $\omega = 2\pi F$ . It is a quantity that measures the rotation rate. Frequency gives the number of cycles over seconds. Angular frequency is expressed as radians per second.
The relation between frequency and time period is given as
$F = \dfrac{1}{T}$
Where, $F$ is the frequency and $T$ is the time period.
The time period is given as
$T = \dfrac{{2\pi }}{\omega }$
Where, \[\omega \] is the angular frequency
Complete step by step answer:
An oscillation is a back and forth motion. In the case of a simple pendulum one oscillation is one complete cycle of swinging and returning back to its original position. The time taken to complete one oscillation is taken as the time period of the pendulum.
We know that frequency of oscillation is the number of oscillations in one second. Frequency is expressed in the unit Hertz. Its symbol is $Hz$. This can be calculated by taking the reciprocal of time period.
The relation between frequency and time period is given as
$F = \dfrac{1}{T}$ (1)
Where, $F$ is the frequency and $T$ is the time period.
We know that time period is inversely related to the angular frequency.
$T = \dfrac{{2\pi }}{\omega }$ (2)
Where, $\omega $ is the angular frequency.
Here, $2\pi $ is in radians. And $\omega $ is measured in radians per second.
Now let us substitute the value of time period as expressed in equation (2) in equation for frequency.
Then we get,
$F = \dfrac{1}{{\dfrac{{2\pi }}{\omega }}}$
$\therefore F = \dfrac{\omega }{{2\pi }}$
This is the relation connecting frequency of oscillation and angular frequency.
So, the correct answer is option C.
Note: From this relation we can see that the angular frequency $\omega = 2\pi F$ . It is a quantity that measures the rotation rate. Frequency gives the number of cycles over seconds. Angular frequency is expressed as radians per second.
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