Question

A sound wave has a propagation speed of 349 m/s, a frequency 1026 Hz, and has a maximum displacement of y_m= 3.02x10^-7 m. At time t= 0, and x= 0, the displacement is y_m = 1.00x10^-7 m.

(a) Using these values, find an equation of form:

           y(x,t) = y_m sin(kx - ωt + ϕ)

(b) Differentiate the expression and find the maximum speed of the oscillating air molecules.

(c) Find the maximum magnitude of acceleration of the oscillating air molecules.

Answer 1

Given that ::

speed of the sound wave, v = 349 m/s

frequency of the sound wave, f = 1026 Hz

maximum displacement, y_m = 3.02 x 10^-7 m

At time t= 0, and x= 0, the displacement is given as, y_m = 1 x 10^-7 m

(a) the equation is given as ::

y (x, t) = y_m Sin (kx - t + )                                  { eq.1 }

where, k = 2 /

or = v / f = ( 349 m/s) / (1026 Hz)

= 0.34 m

k = 2 (3.14) / (0.34 m)

k = 18.4 m^-1

and = 2 f = 2 (3.14) (1026 Hz)

= 6443.2 rad/s

inserting all these values in eq.1,

y (0, 0) = (1 x 10^-7 m) Sin [(18.4)x - (6443.2)t + )          

(b) the maximum speed of the oscillating air molecules is given as ::

v_max = A                                     { eq.2 }

where, A = apmlitude of the wave = 1 x 10^-7 m

inserting the values in eq.2,

v_max = (6443.2 rad/s) (1 x 10^-7 m)

v_max = 6.4 x 10^-4 m/s

(c) the maximum magnitude of acceleration of the oscillating air molecules is given as ::

a_max = ² A                                         { eq.3 }

inserting the values in eq.3,

a_max = (6443.2 rad/s)² (1 x 10^-7 m)

a_max = 41514826.2 x 10^-7 m/s²

a_max = 4.15 m/s²