? Physics2lO Cbp. Ii. sound Waves jrn oit an ambubncc emits a lone deiscnbed by

Question

? Physics2lO Cbp. Ii. sound Waves jrn oit an ambubncc emits a lone deiscnbed by the foflowing equation P?(I.25 wiere P is che preurc fluctuation above cw below t aim. espresaed in pascais (I Pa = I Nfm? z s distanca (in meters). snds is the timc in seconch (I) What is the Wavelength lin m) o1thi sound wave? (2) What h its frequcny (in liz;? (3) What i the speed of these sound Waves (in m?a)? (4) WhasischcpchUr(jnPaz0.5)mM,_O(X)IZs? (5) 11 you ale standing on the sidewalks 1iM frequency would you hear as tus ambulance alipmsehes at a speed of 20 us?s? (6) What frrency would you hear as chi arohulance drove away from you at a speed of 20 mIs? (extra credilt h t a summer N winter day? Expiai..

Explanation / Answer

compare the given equation with standard equation

p = Pmax*sin(k*x - w*t)

1) so, k = 8*pi/3

2*pi/lamda = 8*pi/3

==> lamda = 6/8

= 0.75 m

2) w = 880*pi

f = w/(2*pi)

= 880*pi/(2*pi)

= 440 Hz

3) v = w/k

= (880*pi)/(8*pi/3)

= 330 m/s

4) at x = 0.5 m, t = 0.0012

p = 1.25*sin(8*pi*0.5/3 - 880*pi*0.0012)

= 0.0136 pa

5) Apply Doppler's effect,

f' = f*v/(v - v_source)

= 440*330/(330 - 20)

= 468.4 Hz

6) Apply Doppler's effect,

f' = f*v/(v + v_source)

= 440*330/(330 + 20)

= 414.86 Hz

It must be a winter a day.

because speed of sound wave, v = 330 + 0.6*T

here temperature approximtely zero degrees celsius.

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