(4) A standing sound wave in a tube ( B, ? 0 , length L , cross-sectional area A

Question

(4) A standing sound wave in a tube (B, ?0, length L, cross-sectional area A) is described by s(x,t) = s0 cos(kx)cos(kvt),

where v2 = B/ ?0. (Consider B, ?0, L, A, k, s0, ?x to be known quantities. k here refers to the

wavenumber.)

(a) Find the pressure variation p(x,t) corresponding to s(x,t) and show that pressure nodes are

displacement antinodes and vice versa.
(b) Since any small element of air moves in SHM, the net force on the element must be proportional

to its displacement from equilibrium. Find the effective

Explanation / Answer

a) P(x,t) = BK*So*cos(kx)cos(kvt)

If there is a dispalcement anti node at a particular point, then presuure at that point is low beacuse density at that point is low .hence there is a less pressure than normal , and it creates pressure antinodes ...

similary if there is dispalecment node , then density at that point is more than normal , hence pressure is maxium.

b) F = force =- A*P(x,t)

       = -A*B*K(w/v)*x*So*cos(kx)cos(kvt)

       =-[AB*So*w^2/v^2]*x*cos(kx)*cos(kvt)

for small x , coskx =1

hence   F =-[AB*So*w^2/v^2]*x*cos(kvt)

         spring constant = [AB*So*w^2/v^2]*

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