A `3 m` long organ pipe open at both ends is driven to third harmonic standing wave. If the ampulitude of pressure oscillations is `1` per cent of mean atmospheric pressure `(p_(o) = 10^(5) Nm^(2))`. Find the ampulited of particle displacement and density oscillations. Speed of sound `upsilon = 332 m//s` and density of air `rho = 1.03 kg//m^(3)`.

Answer

A, B, C

Explanation

Given length of pipe, `l = 3 m`
third harmonic implies that
`3 ((lambda)/(2)) = l`
or `lambda = (2l)/(3) = (2 xx 3)/(3) = 2 m`
The angular frequency is
`omega = 2 pi f`.
`= (2 pi nu)/(lambda) = ((2pi)(332))/(2)` or
`omega = 332 pi rad//s`
The partical displacement `y(x, t)` can be written as
`y(x,t) = A cos kx sin omega t`
`k = (2 pi)/(lambda) = (2 pi)/((2l//3)) = (3 pi)/(l)`
and `omega = k nu = (3 pi nu)/(l)` `(nu = (omega)/(k))`
`:. y(x, t) = A cos ((3 pi x)/(l)). sin ((3 pi nu)/(l)) t`
The longitudinal oscillations of an air column can be viwed as oscillations of particle displacement or pressure wave or density wave. Pressure variation is related to particle displacement as
`p(x,t) = - B (del y)/(del x)` (`B =` Bulk modulus)
`= ((3 BA pi)/(l)) sin ((3 pi x)/(l)) sin ((3 pi nu)/(l)) t`
The amplitude of pressure variation is
`p_(max) = (3 BA pi)/(l) rArr = sqrt((B)/(rho))` or `B = rho nu^(2)`
`:. p_(max) = (3 rho nu^(2) A pi)/(l)` or `A = (p_(max)l)/(3 rho nu^(2) pi)`
Here, `p_(max) = 1%` or `p_(0) = 10^(3) N//m^(2)`
Subsiting the values,
`A = ((10^(3))(3))/((3)(1.03)(332)^(2) pi) = 0.0028 m`
or `A = 0.28 cm`
According to the definition of Bulk modulus `(B)`
`B = (- bP)/((dV//V))` ..(i)
Volume `= (Mass)/(Density)` or `V= (m)/(rho)`
or `dV = - (m)/(rho^(2)) drho = - (V d rho)/(rho)`
or `(dV)/(V) = - (drho)/(rho)`
Subsituting in Eq. (i), we get
`d rho = (rho(dP))/(B)`
or ampulitude of density oscillation is
`d rho_(max) = (rho)/(B) p_(max) = (p_(max))/(nu^(2))` `((rho)/(B) = (1)/(nu^(2)))`
`= (10^(3))/(332)^(2) = 9 xx 10^(-3) kg//m^(3)`