Question

If the speed of longitudinal wave equals 10 times the speed of the transverse waves in a stretched wire of material which has modulus of elasticity E, then the stress in the wire is
$\eqalign{
  & A.100E \cr
  & B.\dfrac{E}{{100}} \cr
  & C.\dfrac{E}{{10}} \cr
  & D.10E \cr} $

Answer 1

Hint: Incase of longitudinal waves take an example of the sound wave and we have a formula to find out the speed of sound wave in particular medium having certain elastic modulus(E). Similarly we have a formula to find out what will be the speed of the transverse wave in the wire having tension(T) and linear mass density ($\mu $). Then we equate the speed of the longitudinal wave to ten times the speed of the transverse wave to find out stress in the wire.

Formula used:
$\eqalign{
  & {S_L} = \sqrt {\dfrac{E}{\rho }} \cr
  & {S_T} = \sqrt {\dfrac{T}{\mu }} \cr} $
$\eqalign{
  & \dfrac{T}{A} = stress \cr
  & \dfrac{{\Delta l}}{l} = strain \cr
  & \dfrac{{stress}}{{strain}} = E \cr} $

Complete answer:
Let the speed of longitudinal wave in the stretched wire be S_L and speed of the transverse wave in the stretched wire be S_T
We know that
$\eqalign{
  & {S_L} = \sqrt {\dfrac{E}{\rho }} \cr
  & {S_T} = \sqrt {\dfrac{T}{\mu }} \cr} $
Where
E = elastic modulus or young’s modulus of elasticity
 ρ = volumetric mass density of wire
T = tension in the wire
$\mu $= linear mass density of wire

Now
$\eqalign{
  & \dfrac{T}{A} = stress \cr
  & \dfrac{{\Delta l}}{l} = strain \cr
  & \dfrac{{stress}}{{strain}} = E \cr} $

Where
 A = area of cross section of wire
l = initial length of the wire
$\Delta l$ = variation in length of wire due to tension applied i.e final length – initial length
STRESS/STRAIN = E
STRESS = STRAIN $\times$ E
$\dfrac{T}{A} = \dfrac{{\Delta l}}{l} \times E$ … eq 1
$\eqalign{
  & \rho = \dfrac{\mu }{A} \cr
  & \mu = \rho \times A \cr
  & {S_T} = \sqrt {\dfrac{T}{\mu }} = \sqrt {\dfrac{T}{{\rho \times A}}} = \sqrt {\dfrac{{E \times \dfrac{{\Delta l}}{l}}}{\rho }} = {S_L} \times \sqrt {\dfrac{{\Delta l}}{l}} \cr} $
Since
${S_L} = \sqrt {\dfrac{E}{\rho }} $
It is given in the question that ratio of longitudinal wave speed and transverse wave speed is 10
By equating that and from equation 1 we get
$\eqalign{
  & {S_L} = 10 \times {S_T} \cr
  & \Rightarrow \dfrac{{{S_T}}}{{{S_L}}} = \dfrac{1}{{10}} \cr
  & \Rightarrow \dfrac{{{S_T}}}{{{S_L}}} = \sqrt {\dfrac{{\Delta l}}{l}} \cr
  & \Rightarrow \sqrt {\dfrac{{\Delta l}}{l}} = \dfrac{1}{{10}} \cr
  & \Rightarrow \dfrac{{\Delta l}}{l} = \dfrac{1}{{100}} \cr
  & \Rightarrow \dfrac{T}{A} = \dfrac{1}{{100}} \times E \cr
  & \Rightarrow STRESS = \dfrac{E}{{100}} \cr} $

So, the correct answer is “Option B”.

Note:
The velocities which we consider over here are velocities of waves in particular medium but not velocities of individual particles oscillating in that wave. Velocity of waves is constant as long as material density and source frequency remain constant but particle velocity at every position of wave is different. At maximum displacement position of wave, particle velocity there is minimum and at minimum displacement position particle velocity is maximum.