Question

The opening to a cave is a tall, 30-cm-wide crack. A bat that is preparing to leave the cave emits a 30 kHz ultrasonic chirp. How wide is the "sound beam" 100 m outside the cave opening? Use vsound = 340 m/s.

Answer 1

Width of sound beam is 7.557 m

Step-by-step explanation:

First we will calculate the wave length from given data:

λ=v/f

Were:

v is the speed

f is the frequency

`\lambda=(340)/(30*10^3)\\ \lambda=0.01133 m`

We considered the opening long and narrow, Using single slit diffraction formula:

mλ=dsinΘ

where:

d is the crack width

m is the order

Θ is angle

Considering m=1, The angle between first minimum from center of beam is:

`\theta=sin^(-1)((m\lambda)/(d))\\\theta=sin^(-1)((1*0.01133)/(30*10^(-2)))\\ \theta=2.164^o`

The width of beam is:

tanΘ=y/L

`tan\theta=(w/2)/(L)\\ w=2L\ tan\theta\\w=2*100*tan 2.164\\w=7.557 m`

Width=7.557 m