Question

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

Answer 1

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 - $v_{s}$ ) / ( v + $v_{0}$ ) ] × f

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