Question

tudent A runs down the hallway of the school at a speed of vo=5.00m/s, carrying a ringing 1024.00-Hz tuning fork toward a concrete wall. The speed of sound is v=343.00m/s. Student B stands at rest at the wall. (a) What is the frequency heard by student B? (b) What is the beat frequency heard by student A?

Answer 1

Using the Doppler Effect, we can calculate that student B will hear a frequency of 1038.82 Hz from student A's tuning fork. The beat frequency heard by student A, caused by the interference of the original and reflected sounds, will be 14.82 Hz.

Step-by-step explanation:

For student A running towards the wall with a 1024 Hz tuning fork, the frequency heard by student B can be found using the Doppler Effect equation for approaching sources:

f' = f * (v + v0) / v

Where f' is the frequency heard by the observer, f is the original frequency of the tuning fork, v is the speed of sound, and v0 is the speed of the source towards the observer. Substituting the given values:

f' = 1024 Hz * (343 m/s + 5 m/s) / 343 m/s

f' = 1024 Hz * 1.0146 = 1038.82 Hz

The beat frequency heard by student A is the difference between the frequency of the reflected sound from the wall and the frequency of the tuning fork when both frequencies mix:

Δf = |f' - f|

Δf = |1038.82 Hz - 1024 Hz| = 14.82 Hz

Answer 2

a)
`f'=1039.15\ Hz`

b)
`f_b=15.15\ Hz`

Step-by-step explanation:

Given:

velocity of student A carrying the sound source,
`v_s=5\ m.s^(-1)`

frequency of sound,
`f=1024\ Hz`

speed of sound,
`v=343\ m.s^(-1)`

(a)

Mathematical expression of Doppler-Effect is given as:

`f'=((v+v_o)/(v-v_s) )f`

where:

f' = observed frequency

f = frequency of sound

`v_o=`velocity of the observer

Putting respective values in the above eq.:

`f'=((343+0)/(343-5) )* 1024`

`f'=1039.15\ Hz`

(b)

Beat frequency will be the frequency difference between the sound source and its reflection from the walls.

`f_b=f'-f`

`f_b=1039.15-1024`

`f_b=15.15\ Hz`