Question

A bungee jumper jumps from a tall bridge that is 60m above a flat, dry creek bed. She is carrying an audio generator that emits sound at frequency 1200 Hz. The unstretched length of the bungee cord is 26m, and the spring constant is selected so that she stops just before hitting the creek bed. At what point in her fall is the beat frequency, determined by listening to the audio generator and the reflection from the creek bed, a maximum

Answer 1

Answer:

Explanation:

The original frequency of sound  f₀

The apparent frequency of sound fa

For apparent frequency the formula is

fa =  $f_0\times\frac{V+v}{V-v }$     , v is velocity of jumper which increases as he goes down .

Beat frequency

= fa - f₀

=  $f_0\times(\frac{V+v}{V-v }-1)$

=  $f_0\times(\frac{2v}{V-v })$

since v is very small in comparison to V , velocity of sound , in the denominator , v can be neglected.

beat frequency = $f_0\times(\frac{2v}{V })$

v , the velocity of jumper will go on increasing as long as net force on the jumper is positive or

mg > kx where x is extension in the cord and k is its force constant . Below this point kx or restoring force becomes more than weight of the jumper and then net force on the jumper directs upwards. At this point beat frequency becomes maximum.