deduce the expression for the speed of a compressional wave through an extended solid
Answers
Answered by
0
A mechanical longitudinal wave such as Sound wave which is propagated by the elastic compression of medium is called As compressional wave.
According to me there is no extended solid . I think it should be Only Solid.
Expression For Speed of Compressional wave in Solid:
When a thin rod of large length is stretched in the direction of its length, there will be increase in length with contraction in cross section and rod is set in longitudinal vibration.
Let ρ be density.
Υ be young's modulus of the material of the rod.
Let α be the cross sectional of rod.
Consider an elementary slice AB in the rod bounded by two planes A and B at x and x+dx respectively at right angles .
Let these planes be displaced by y and y+dy amount.
The distance between planes A,B :(x+dx+y+dy)-(x+y)=dx+dy
so increase in length is dy and increase in length per unit length is dy/dx
If f is total force acting on area α then
longitudinal stress=f/α
Young's modulus Y=f/α/dy/dx
f=Υαdy/dx-----------------(1)
the fore acting on second face is f+(df/dx)dx---------2
substututing value of f in equation 2
we get:
Υαdy/dx +(d/dx(Υαdy/dx))dx --------------3
resultant fore is given by difference of expression 3 and 1
Υαdy/dx +(d/dx(Υαdy/dx))dx -Υαdy/dx
=αY( d²y/dx²) dx
but mass of element dx is ραdx acceleration d²y/dt²
hence
d²y/dt²=Y/ρ( d²y/dx²)
d²y/dt²=c²d²y/dx²-----------(4)
where c²=Y/ρ where c is the velocity of wave propagation.from equation 4 which is a wave equation, so its solution should of the form y=f(ct-x)+F(ct+x)
Hence we get longitudinal wave velocity as c=√Y/ρ
According to me there is no extended solid . I think it should be Only Solid.
Expression For Speed of Compressional wave in Solid:
When a thin rod of large length is stretched in the direction of its length, there will be increase in length with contraction in cross section and rod is set in longitudinal vibration.
Let ρ be density.
Υ be young's modulus of the material of the rod.
Let α be the cross sectional of rod.
Consider an elementary slice AB in the rod bounded by two planes A and B at x and x+dx respectively at right angles .
Let these planes be displaced by y and y+dy amount.
The distance between planes A,B :(x+dx+y+dy)-(x+y)=dx+dy
so increase in length is dy and increase in length per unit length is dy/dx
If f is total force acting on area α then
longitudinal stress=f/α
Young's modulus Y=f/α/dy/dx
f=Υαdy/dx-----------------(1)
the fore acting on second face is f+(df/dx)dx---------2
substututing value of f in equation 2
we get:
Υαdy/dx +(d/dx(Υαdy/dx))dx --------------3
resultant fore is given by difference of expression 3 and 1
Υαdy/dx +(d/dx(Υαdy/dx))dx -Υαdy/dx
=αY( d²y/dx²) dx
but mass of element dx is ραdx acceleration d²y/dt²
hence
d²y/dt²=Y/ρ( d²y/dx²)
d²y/dt²=c²d²y/dx²-----------(4)
where c²=Y/ρ where c is the velocity of wave propagation.from equation 4 which is a wave equation, so its solution should of the form y=f(ct-x)+F(ct+x)
Hence we get longitudinal wave velocity as c=√Y/ρ
Similar questions