Question

You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they begin to let out a yell of excitement (or possibly fear) that sounds like the musical note G (392 Hz) as they fall towards you and E (330 Hz) as they bounce back away from you. They continue to yell as they bounce up and down in simple harmonic motion. You notice the pitch of their yell changes as they move toward and away from you. The time in between the highest pitch and lowest pitch is 10 seconds. What is the amplitude of their simple harmonic motion

Answer 1

The amplitude is
`A = 90.2 \ m`

Step-by-step explanation:

From the question we are told that

The frequency of when sound is approaching observer is
`f = 392 Hz`

The frequency as the move away from observer is
`f_ a = 330 \ Hz`

The time between the pitch are
`t = 10 \ s`

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

`T = 2 t`

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

`T = (2 \pi)/(w)`

Where
`w` is the angular velocity

=>
`(2 \pi)/(w) = 2 * 10`

=>
`w = 0.314 \ rad/sec`

Now using Doppler Effect,

The source of the sound is approaching the observer

The

`f = f_o ((v)/(v- wA) )`

`392 = f_o ((v)/(v- wA) )`

Where A is the amplitude

So when the source is moving away from the observer

`f_a = f_o ((v)/(v+ wA) )`

`330 = f_o ((v)/(v+ wA) )`

Here
`f_o` is the fundamental frequency

Dividing the both equation we have

`(392)/(330) = (f_o((v)/(v-wA) ))/(f_o((v)/(v+wA))`

`1.1878 = (v+wA)/(v-wA)`

`1.1878 v - 1.1878 wA = v+wA`

`1.1878 v = 2.1878 wA`

=>
`A = ((0.1878 * (330)))/((2.1878)* (0.314))`

`A = 90.2 \ m`