Question

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How wide is the "sound beam" 100. M outside the cave opening? Use v sound= 340. M/s.

Answer 1

Answer:

The value is  $w = 7.54 \ m$        

Explanation:

From the question we are told that

     The length of the crack is  $a = 0.3 \ m$

     The  frequency is  $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

      The distance outside the cave that is being consider is  $D = 100 \ m$

      The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

        $\lambda = \frac{v}f}$

=>     $\lambda = \frac{340 }{30*10^{3}}$

=>     $\lambda = 0.0113 \ m/s$

Generally for a  single slit the path difference between the interference patterns of the sound wave and the center  is mathematically represented as  

          $y = \frac{ n * \lambda * D}{a}$

=>     $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=>     $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

         $w = 2 * y$

=>      $w = 2 * 3.77$

=>      $w = 7.54 \ m$