contestada

Two loudspeakers, A and B, are driven by the same amplifierand
emit sinusoidal waves in phase. The frequency emitted byeach
speaker is 172 Hz. You are 8.00 m from A. What isthe closest you
can be to B and be at a point of destructiveinterference.

Respuesta :

Answer:

1 m

Explanation:

given,

frequency emitted by the speaker,

f = 172  Hz

speed of sound,v = 344 m/s

distance of Point A = 8 m

[tex]wavelength = \dfrac{speed\ of\ sound}{frequency}[/tex]

[tex]\lambda= \dfrac{344}{172}[/tex]

   λ = 2 m

For destructive interference

[tex]r_a - r_b = \dfrac{\lambda}{2}[/tex]

[tex] r_b =r_a - \dfrac{\lambda}{2}[/tex]

[tex] r_b = 8 - \dfrac{2}{2}[/tex]

 [tex] r_b =7 m [/tex]

the closest you  can be to B and be at a point of destructive interference  

 = r_a - r_b

 = 8 - 7

  = 1 m

   

Otras preguntas