Question

A string is under tension so that its length is increased by $\dfrac{1}{n}$ times its original length. The ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will be?
A. $1:n$
B. ${n^2}:1$
C. $\sqrt n :1$
D. $n:1$

Answer 1

Hint: In the case of a problem based on reflection of waves phenomena, we know that the frequency of vibrations(waves) directly varies with velocity hence, first we will find the ratio of the velocity of both the longitudinal and transverse waves and then establish a relation between them to find the required ratio.

Formula used:
The Velocity of longitudinal waves is given as,
${v_1} = \sqrt {\dfrac{Y}{\rho }} $
where, Y = Young’s modulus
$\rho $ = density of string
The velocity of transverse waves is given as,
${v_2} = \sqrt {\dfrac{T}{m}} $
where, T = Tension in the string
m = mass per unit length

Complete step by step solution:
According to question, A string is under tension such that its length is increased by $\dfrac{1}{n}$ times its original length i.e., $\Delta l = \dfrac{l}{n}$ (given)
As we know that, the frequency is directly proportional to the velocity, therefore
$\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{{v_1}}}{{{v_2}}}$ … (1)

Now, the formula for the Velocity of longitudinal waves is:
${v_1} = \sqrt {\dfrac{Y}{\rho }} $ … (2)
And, the velocity of transverse waves is:
${v_2} = \sqrt {\dfrac{T}{m}} $
But, mass per unit length is,
$m = \dfrac{{\text{Volume} \times \text{Density}}}{\text{length}} = \dfrac{{A \times l \times \rho }}{l} = A\rho $
${v_2} = \sqrt {\dfrac{T}{{A\rho }}} $ … (3)

Divide eq. (2) by eq. (3), we get
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\sqrt {\dfrac{Y}{\rho }} }}{{\sqrt {\dfrac{T}{{A\rho }}} }} = \sqrt {\dfrac{{YA}}{T}} $ … (4)
Since, Young’s modulus can be given as:
$Y = \dfrac{\dfrac{T}{A}}{\dfrac{\Delta l}{l}} \\ $
$\Rightarrow Y = \dfrac{{Tl}}{{A\Delta l}} \\ $
$\Rightarrow T = \dfrac{{YA\Delta l}}{l} \\ $
$\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{YA}}{{\dfrac{{YA\Delta l}}{l}}}} = \sqrt {\dfrac{l}{{\Delta l}}} = \sqrt n \\ $
$(\Rightarrow \dfrac{{\Delta l}}{l} = \dfrac{1}{n}) \\ $
From equation (1)
$\therefore \dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{{v_1}}}{{{v_2}}} = \sqrt n $
Thus, the ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will be $\sqrt n :1$.

Hence, the correct option is C.

Note: Since this is a problem related to transverse and longitudinal vibration in wave analytics hence, given conditions are to be analyzed very carefully and parameters that are required to calculate the required ratio of frequency must be identified on a prior basis to get the accurate solution of the problem.