Question

The pressure variation in a sound wave in air is given by
$\Delta P=12sin(8.18x-2700t+\pi /4)N/m^2$
Find the displacement amplitude. Density of air = $1.29kg/m^3$
Give answer in terms of 10${}^{-5}$ m

• A. 1.05
• B. 2
• C. 5
• D. 15

Answer 1

Answer: (A)

1.05

Given - $\Delta P=12\sin (8.18x-2700t+\pi /4)$ ,

comparing this equation with , $\Delta P=\Delta p_m\sin (kx-\omega t+\varphi )$ ,

we get , $\Delta p_m=12Pa$ ,

$\omega =2700$ ,

or $2\pi f=2700$ ,

or $f=2700/2\pi$

and $k=8.18$ ,

or $2\pi /\lambda =8.18$ ,

or $\lambda =2\pi /8.18$ ,

therefore $v=f\lambda =\frac{2700}{2\pi }.\frac{2\pi }{8.18}=330m/s$

The relation between pressure amplitude $\Delta p_m$ and displacement amplitude $A$ (maximum value of displacement ) is given by ,

$\Delta p_m=\left(vd\omega \right)A$ ,

where $v=$ speed of sound in air ,

$d=$ density of medium (air) ,

$\omega =$ angular frequency ,

given $v=330m/s,d=1.29kg/m^3,\omega =2700rad/s,\Delta p_m=12Pa$ ,

now $\Delta p_m=\left(vd\omega \right)A$ ,

or $A=\frac{\Delta p_m}{vd\omega }=\frac{12}{330\times 1.29\times 2700}=1.04\times {10}^{-5}m$