Question

A light pointer fixed to one prong of a tuning fork touches a vertical plate. The fork is set vibrating and the plate is allowed to fall freely. Eight complete oscillations are counted when the plate falls through $10cm$ then the frequency of the fork is $(g = 9.8m/{s^2})$
A.$65Hz$
B.$56Hz$
C.$46Hz$
D.$64Hz$

Answer 1

Here, we have to use the second equation of motion to find the relation between the distance covered and time. As the whole arrangement is in the vertical direction, the acceleration is due to gravity only. With the help of the second equation of motion, we have to find the time taken to complete the given distance and hence find the frequency.
Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
Where,
$s$ is the distance covered by the plate.
$u$ is the initial velocity.
$t$ is the time taken.
$a$ is the acceleration due to gravity.

Complete answer:
Let us assume that the plate is at rest initially. The plate covers a distance of $10cm$, i.e. $0.1m$. Now, we have to use the second equation of motion:
$s = ut + \dfrac{1}{2}a{t^2}......(1)$
Where,
$s$ is the distance covered by the plate.
$u$ is the initial velocity.
$t$ is the time taken.
$a$ is the acceleration due to gravity.
As the plate is initially at rest, therefore the initial velocity is zero and acceleration is due to gravity as the plate is falling freely. So, the given quantities are:
$u = 0$
$s = 0.1m$
$a = 9.8m/{s^2}$
We have to put all these values in equation (1) to get,
$s = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow 0.1 = 0 \times t + \dfrac{1}{2} \times 9.8 \times {t^2}$
$\Rightarrow 0.1 = \dfrac{1}{2} \times 9.8 \times {t^2}$
$\Rightarrow \dfrac{{0.1 \times 2}}{{9.8}} = {t^2}$
$\Rightarrow t = \sqrt {\dfrac{{0.1 \times 2}}{{9.8}}}$
$\Rightarrow t = \sqrt {\dfrac{1}{{49}}}$
$\Rightarrow t = \dfrac{1}{7}s$
Now, during this time period, eight oscillations are completed. So, the time taken to complete one oscillation is given by:
Time is taken to complete $8$ oscillations$= \dfrac{1}{7}s$
Time is taken to complete $1$ oscillation$= \dfrac{1}{{7 \times 8}}s$
Time is taken to complete $1$ oscillation$= \dfrac{1}{{56}}s$
But we have to find the frequency which is just the reciprocal of the time period. Mathematically,
$\nu = \dfrac{1}{T}$
Where,
$\nu$ is the frequency.
$T$ is the time period.
$\Rightarrow \nu = \dfrac{1}{{\left( {\dfrac{1}{{56}}} \right)}}$
$\Rightarrow \nu = 56Hz$

Therefore, option (B) is correct.

Note:
The wavelength, frequency, time period, amplitude, and speed are five characteristic features of any type of wave. But we mostly sense two types of waves: sound waves and light waves. Frequency has its features in these two types of waves. In sound waves, frequency is perceived as pitch. The higher the frequency of the waves, the higher is the pitch of the sound.
The range of different frequencies of electromagnetic radiation is known as the electromagnetic spectrum. The light which is visible to us lies in the range of $400THz$ to $790THz$. The gamma rays have the highest frequency and lowest wavelength in the electromagnetic spectrum.