Question

A uniform string of length L and mass M is fixed at both ends while it is subject to a tension T. It can vibrate frequencies $v$ given by the formula (where n=1, 2, 3, ) :
A) $v = \dfrac{n}{2}\sqrt {\dfrac{T}{{ML}}}$.
B) $v = \dfrac{n}{{2L}}\sqrt {\dfrac{T}{M}}$.
C) $v = \dfrac{1}{{2n}}\sqrt {\dfrac{T}{{ML}}}$.
D) $v = \dfrac{n}{2}\sqrt {\dfrac{{TL}}{M}}$.

Answer 1

Frequency of oscillation is defined as the number of oscillations made by an oscillating body in one second. The linear mass density is defined as the ratio of any characteristics value of the body upon the length of the body.

Formula used: The velocity of a wave in the string of length L and tension T is given by,
$V = \sqrt {\dfrac{T}{m}}$
Where T is tension in string and m is the linear mass density of the string.
The length of the string is given by,
$L = \dfrac{{n\lambda }}{2}$
Where L is the length $\lambda$ is the wavelength and n is the multiple.
Also for wave,
$V = \lambda v$
V is the velocity of wave $\lambda$ is the wavelength and f is the frequency of the wave.

Complete step-by-step answer:
It is given in the problem that there is a uniform string of length L and mass M which is fixed at both ends and it has a tension T in the string and we need to find the formula of the frequency of the wave in terms of tension, mass and length.
The length of the string can be expressed in terms of wavelength as,
$L = \dfrac{{n\lambda }}{2}$
Where L is the length $\lambda$ is the wavelength and n is the multiple.
The wavelength can be expressed as,
………eq. (1)
The frequency of the wave can be expressed as,
$\Rightarrow v = \dfrac{V}{\lambda }$………eq. (2)
As the velocity of the wave in the string is given by,
$V = \sqrt {\dfrac{T}{m}}$
Where T is tension in the string and m is the linear mass density of the string.
$\Rightarrow V = \sqrt {\dfrac{T}{m}}$………eq. (3)
Replace the value of velocity V and wavelength $\lambda$from equation (1) and equation (3) in equation (2) we get,
$\Rightarrow v = \dfrac{V}{\lambda }$
$\Rightarrow v = \dfrac{{\left( {V = \sqrt {\dfrac{T}{m}} } \right)}}{{\left( {\dfrac{{2L}}{n}} \right)}}$
$\Rightarrow v = \dfrac{n}{{2L}}\left( {\sqrt {\dfrac{T}{m}} } \right)$
Since $m = \dfrac{M}{L}$
$\Rightarrow v = \dfrac{n}{2}\left( {\sqrt {\dfrac{T}{{M \cdot L}}} } \right)$
So the frequency of the wave is$v = \dfrac{n}{2}\left( {\sqrt {\dfrac{T}{{M \cdot L}}} } \right)$.

The correct answer is option A.

Note: Linear mass density is the ratio of total mass of the body and total length of the body. The length of string which is vibrating with a frequency can be expressed as the product of integral multiples of $\dfrac{\lambda }{2}$. A wavelength is the distance between the consecutive crests of the wave or the distance between consecutive troughs.