A person jumps out of a helicopter in mid-air to retrieve a small metal ball. He screams at 475Hz, but people on the ground hear him scream at 578Hz.
The person is falling at __(m/s)
Once he grabs the metal ball, he is pulled back up to the helicopter at the same speed as he was falling.
People on the ground now hear him scream at __(Hz)
Take the speed of sound to be 343m/s.
A. Fg = (Vs+Vg)/(Vs-Vp) * Fp.
578 = (343+0)/(343-Vp) * 475.
578 = 343/(343-Vp) * 475,
Divide both sides by 475:
1.21 = 343/(343-Vp),
416-1.21Vp = 343, Vp = 60.3 m/s.
B. Fg = (Vs-Vg)/(Vs+Vp) * Fp.
Fg = (343-0)/(343+60.3) * 475.
Since the distance between the people on the gnd. and the person is now increasing, the freq. heard(Fg) is less than 475
Hz.
To solve this problem, we need to use the Doppler effect formula:
f' = (v + vs) / (v + vd) * f
Where:
f' = observed frequency
v = speed of sound = 343 m/s
vs = velocity of the source (person falling)
vd = velocity of the detector (person on the ground)
f = original frequency
First, let's calculate the velocity of the person falling:
Given:
Observed frequency, f' = 578 Hz
Original frequency, f = 475 Hz
Speed of sound, v = 343 m/s
Using the formula, we can rearrange it to solve for vs:
vs = (f' - (v + vd) / (v + vd)) * v
Substituting the given values:
vs = (578 - 343) / (343 + 343) * 343
vs = 235 / 686 * 343
vs ≈ 117.79 m/s
So, the person is falling at approximately 117.79 m/s.
Next, let's calculate the frequency observed on the ground when the person is pulled back up:
Given:
Velocity of the person falling, vs = 117.79 m/s
Velocity of the detector (person on the ground), vd = 0 m/s
Original frequency, f = 475 Hz
Speed of sound, v = 343 m/s
Using the same formula, we can now solve for the observed frequency:
f' = (v + vs) / (v + vd) * f
Substituting the given values:
f' = (343 + 117.79) / (343 + 0) * 475
f' = 460.79 / 343 * 475
f' ≈ 639.24 Hz
Therefore, people on the ground now hear the person scream at approximately 639.24 Hz when he is pulled back up to the helicopter.
To solve this problem, we need to use the concept of the Doppler effect. The Doppler effect describes the change in frequency of a wave (in this case, sound) due to relative motion between the source of the wave and the observer.
First, let's find the speed at which the person is falling. The Doppler effect formula for sound is given by:
f_observed = f_source * (v_sound + v_observer) / (v_sound + v_source)
Where:
- f_observed is the frequency observed by the observer
- f_source is the frequency emitted by the source
- v_sound is the speed of sound
- v_observer is the speed of the observer (in this case, the person falling)
- v_source is the speed of the source (in this case, the metal ball)
Given:
- f_observed = 578 Hz
- f_source = 475 Hz
- v_sound = 343 m/s
Rearranging the formula:
v_observer = (f_observed / f_source) * (v_sound + v_source) - v_sound
Plugging in the values:
v_observer = (578 / 475) * (343 + 0) - 343
v_observer = 208.42 m/s (rounded to 2 decimal places)
Therefore, the person is falling at a speed of approximately 208.42 m/s.
Now, let's find the frequency observed when the person is pulled back up to the helicopter at the same speed.
Using the same Doppler effect formula as before, but now considering the person as the observer (v_observer = 0), we can calculate the the frequency observed:
f_observed = f_source * (v_sound + v_observer) / (v_sound + v_source)
Given:
- f_source = 578 Hz
- v_sound = 343 m/s
- v_observer = 0
Plugging in the values:
f_observed = 475 * (343 + 0) / (343 + 208.42)
f_observed = 319.66 Hz (rounded to 2 decimal places)
Therefore, when the person is pulled back up to the helicopter, people on the ground will hear him scream at approximately 319.66 Hz.