Question

A uniform wire of length L, diameter D and density $\rho $ is stretched under a tension T. The correct relation between its fundamental frequency f, the length L and the diameter D is
a. $f \propto \dfrac{1}{{LD}}$
b. $f \propto \dfrac{1}{{L\sqrt D }}$
c. $f \propto \dfrac{1}{{{D^2}}}$
d. $f \propto \dfrac{1}{{L{D^2}}}$

Answer 1

Hint: The Frequency for a wave travelling in a medium is directly proportional to the reciprocal of its wavelength. The fundamental frequency is known to be such, the wavelength is of twice the length of the string. Rearrange the mathematical equations for these concepts to find the correct option among the given ones.
Formula Used: $f = \dfrac{1}{{2L}} \times \sqrt {\dfrac{T}{\mu }} $

Complete answer:
It is given that, frequency is ‘f’, diameter is ‘D’, length is ‘L’, density is ‘$\rho $’, and tension is ‘T’.
We know that Frequency is defined as the number of complete vibration cycles of a medium in one second. It covers a distance of its wavelength in one cycle. Its unit is hertz.
Mathematically it is given as, $f = \dfrac{c}{\lambda }$
Where ‘f’ is the frequency which is expressed in hertz, c’ is the speed of the wave in a medium it is expressed in terms of m/s and ‘$\lambda $’ is the wavelength of the wave it is expressed in metre.
The fundamental vibrational mode of stretched string is known as that the wavelength is of twice the length of the string. It is seen that resonant standing wave modes are produced in a stretched string.
Mathematically it is given as,
$f = \dfrac{1}{{2L}} \times \sqrt {\dfrac{T}{\mu }} $
Where ‘f’ is the frequency, ‘L’ is the length of the given uniform wire, ‘T’ is the tension in wire, and ‘$\mu $’ is the mass per unit length
$ \Rightarrow f = \dfrac{1}{{2L}} \times \sqrt {\dfrac{T}{\mu }} $
Where ‘$\mu $’ is the mass per unit length which is given as
$\mu = \pi \times \dfrac{{{D^2}}}{{4\rho }}$
Now, relation between its fundamental frequency f, the length L and the diameter D is
is given as,
$f = \dfrac{1}{{2L}} \times \sqrt {\dfrac{T}{{\pi \times \dfrac{{{D^2}}}{{4\rho }}}}} $
Since, ‘T’, ‘$\pi $’ and ‘$\rho $’ are constants
$f \propto \dfrac{1}{{LD}}$

Hence, (a) is the correct option.

Note:
Waves that need a medium for their propagation are known as mechanical waves or elastic waves. When the wave propagates through the medium the particles of the medium execute periodic motion about its mean position. Waves which do not need a medium for its propagation are known as the non - mechanical waves, they are transverse in nature.