Question

The time period of a tuning fork is one thousandth of a second. This means that
A. The tuning fork vibrates $1$ time in $1000$ seconds
B. The tuning fork vibrates $10$ times in $100$ seconds
C. The tuning fork vibrates $1000$ times in $1$ second
D. The tuning fork vibrates $100$ times in $10$ seconds

Answer 1

Hint:The tuning fork vibrates in an oscillatory motion exhibiting simple harmonic motion. Hence the frequency of these oscillations are obtained by applying the formula for the frequency with respect to its time period since frequency and time period are inversely related to each other.

Complete step by step answer:
Tuning fork is an example of simple harmonic motion as this motion is also a periodic motion. This is because the mechanism of a tuning fork is such that when struck on a hard object the metal prongs start to vibrate about its mean position, that is, the original position of the forks. This vibratory motion is periodic which means that the vibrations occur at regular intervals. The particles suffer a displacement from its mean position and hence it executes simple harmonic motion.

This means that there will be a certain time period involved with these oscillations since it is a periodic motion. The time period is defined as the time taken by the particle to complete a single oscillation. Since it is a periodic motion that repeats constantly we can say that this time period is said to be the smallest time after which the oscillation repeats.

The oscillation here specifies the to and fro motion of the particles of the prongs which start and ends at the same position. The oscillation is nothing but the vibrations in a tuning fork.We are asked to find the frequency of these oscillations. The frequency is said to be the number of oscillations per unit time. It is directly related to the time period.

Since these quantities are seem to be the reciprocal of one another, the equation relating them is given by:
$f = \dfrac{1}{T}$ ---($1$)
Where, $f$ is the frequency of the oscillations and $T$ is the time period of the oscillations.
Here, the time period is given to be one thousandth of a second which means that:
$T = \dfrac{1}{{1000}}\,{s^{ - 1}}$
Hence we substitute this value into the equation ($1$) and we get:
$f = \dfrac{1}{{\dfrac{1}{{1000}}}}$
$ \therefore f = 1000\,{s^{ - 1}}$
Hence we can see that the oscillation of the tuning fork occurs $1000$ times and by definition these oscillations occur per unit time which means per second. This means that when the time period of the tuning fork is one thousandth of a second then its corresponding frequency is said to be $1000$ times per second.

Hence the correct answer is option B.

Additional information: Tuning forks are made up of a metal namely Aluminium are used to produce vibrations with a constant frequency. They are hence used in many applications like in the medical field to test the extent of hearing loss in a patient and this test is known as the Rinne test. Usually the tuning fork is used in musical instruments to match the wave pattern of an instrument hence it is used as an acoustic resonator. Other applications include X-rays and sound therapy.

Note:The vibratory motion of the tuning fork only occurs when its forks are struck onto a surface (usually a rubber pad), that is, only when an external force is applied the oscillatory motion occurs. The motion not only oscillates but also produces a sound as the particles in air vibrate and are transmitted in the form of waves of sound.