The figure shows the output from a pressure monitor mounted at a point along the

Question

The figure shows the output from a pressure monitor mounted at a point along the path taken by la sound wave of a single frequency travelling at 344 m/s through air with a uniform density of 1.51 kg/m^3. The vertical axis scale is set by deltap_s = 4.40 mPa. If the displacement function of the wave is written as s(x, t) = s_m, cos(kx - omegat), what are (a) s_m, (b) k, and (c) omega? The air is then cooled so that its density is 1.95 kg/m^3 and the speed of a sound wave through it is 323 m/s. The sound source again emits the sound wave at the same frequency and same pressure amplitude. What now are (d) s_m, (e) k, and (f) omega?

Explanation / Answer

We have the pressure wave given by: p = pm sin( kx - t) which gives rise to the displacement wave given by:

s = sm cos(kx - t)   

With ps = 4.4 mPa; density () = 1.51 Kg/m3  and speed of sound (v) = 344 m/s

Moreover from the graph we see that time period of oscillation (T) = 2.3 s

(c) Hence the frequency of sound wave is () = 2/T = 2*3.14/2.3 = 2.73 Hz

(b) Again the wave no. k is given by k = /v = 2.73/344 = 0.0079 m-1

(a) Also we have the relation pm = *v**sm   hence sm = 4.4*10-3/1.51*344*2.73 = 3.1*10-6 m

(d) Now we have got = 1.95 Kg/m3 and v = 323 m/s Hence sm = 4.4*10-3/1.95*323*2.73 = 2.56*10-6 m

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