Question

A whistle of frequency fo = 1300Hz is dropped
from a height H=505 m above the ground. At
the same time, a detector is projected
upwards with velocity u = 50 ms' along the
same line. If the velocity of sound is
V=300ms 1, The frequency detected by the
detector after t = 5s is (g=10ms 2)​

Answer 1

Answer:

Given

f° = 1300Hz

H = 505

Let t1 be the time at which source emits sounds at t= 5 secs

then , the time duration for which so and move is t-t1..h= ut-½gt²

H = ½g(t1)²-h=v(t-t1)

505 -½g(t1)-ut +½gt²=v(t-t1)

505-5(t1)²-125 =300 (5-t1)

(t1)²-60(t1)+224=0

t1=4 sec

f=f°{v+V1/v-v1}

v°= v-gt

V1=gt= 40 ms-1

= 1300×300/(300-40)

= 1500Hz

Answer 2

Answer:

The frequency detected by the detector after t = 5s is 1500HZ

Explanation:

A whistle of frequency fo = 1300Hz

A whistle of frequency fo = 1300Hz is dropped from a height H=505 m above the ground.

At the same time, a detector is projected upwards with velocity u = 50 ms' along the same line.

If the velocity of sound is $V=300ms^{-1}$

Time take to detect the sound (t)=5sec

Consider acceleration due to gravity (g) = $10ms^{2}$

$h=ut+\frac{1}{2} gt^{2}$

and

$H-\frac{1}{2} gt^{'2} -h=v(t-t')\\505=\frac{1}{2} gt'^{2} -(ut-\frac{1}{2} gt^{2} )=v(t-t')$

Substituting the values  t=5s, g=10$ms^{-1}$, and v= $300 ms^{-1}$

$505-5t'^{2} -125=300(5t')\\t'^{2} -60t'+224=0\\t'=4s$

By the Dopplers formula  $f_{app} =f_{0} \frac{v+v_{0} }{v-v_{s} }$

      where $v_{s} =gt'=10*4=40ms^{-1}$

         $v_{0} =u-gt=50-10*5=0$

       $f_{app} =1300 \frac{300+0 }{300-40 }\\\\=1500 HZ$

The frequency detected by the detector after t = 5s is 1500HZ

The project code is #SPJ2