contestada

An alarm clock is dropped off the edge of a tall building. You, standing directly under it, hear a tone of 1350 Hz coming from the clock at the instant it hits the ground. Since you know the building is 25 m tall, you can find out what the frequency of the alarm would be if you had just held it in your hands. What would that frequency be

Respuesta :

nuhulawal20

Answer:

the frequency clock would be 1262.85 Hz

Explanation:

Given data;

height of building h = 25 m

from the third equation of motion;

v² = u² + 2as

Since the Alarm clock falls with an acceleration equal to the acceleration due to gravity; a = g = 9.81 m/s²

initial velocity u = 0

so we substitute our values into the kinematic equation

v² = (0)² + 2 × 9.81 × 25

v² = 490.5

v = √490.5

v = 22.1472 m/s

Now, since the alarm clock is moving both I am stationary;

my velocity will be zero.

so Frequency of the alarm clock will be;

f' = [ (v - [tex]v_{s}[/tex] ) / ( v + [tex]v_{0}[/tex] ) ] × f

we know that; speed of sound is 343 m/s, so v = 343 m/s, [tex]v_{s}[/tex] is 22.1472 m/s, f is 1350 Hz, [tex]v_{0}[/tex]  is 0 m/s

so we substitute the values into the equation

f' = [ (343 - 22.142 ) / ( 343 + 0 ) ] × 1350

f' = [ 320.858 / 343 ] × 1350

f' = 0.935446 × 1350

f' = 1262.85 Hz

Therefore, the frequency clock would be 1262.85 Hz

onyebuchinnaji

The source frequency or frequency of the alarm is 1,262.25 Hz.

The given parameters:

  • Observed frequency, Fo = 1350 Hz
  • Height of the building, h = 25 m
  • Speed of sound, V = 343 m/s

The source velocity is calculated as follows;

[tex]v^2 = u^2 + 2gh\\\\ v^2 = 0 + 2gh\\\\ v = \sqrt{2gh} \\\\ v = \sqrt{2 \times 9.8 \times 25} \\\\ v = 22.14 \ m/s[/tex]

The source frequency or frequency of the alarm is calculated by applying Doppler effect as follows;

[tex]f_s = f_o (\frac{v- v_s}{v+ v_0} )\\\\ f_s = 1350 (\frac{343-22.14}{343} )\\\\ f_s = 1,262.25 \ Hz[/tex]

Learn more about Doppler effect here: https://brainly.com/question/3841958

Otras preguntas