Question

A loudspeaker at the base of a cliff emits a pure tone of frequency 3000.0 hz. a man jumps from rest from the top of the cliff and safely falls into a net below. how far has the man fallen at the instant he hears the frequency of the tone as 3218.0 hz? the speed of sound is 343 m/s.

Answer 1

Using the Doppler Effect formula, the man's velocity during descent was found to be approximately 26.7 m/s. With this velocity, the time taken to reach that speed was calculated to be around 2.73 seconds. Using the time in the free fall distance equation, the man has fallen approximately 36.4 meters.

Step-by-step explanation:

To determine how far the man has fallen at the instant he hears the frequency of the tone as 3218.0 Hz, we must use the Doppler Effect formula:

f' = f (v + vo) / (v + vs)

Where f' is the observed frequency (3218.0 Hz), f is the emitted frequency (3000.0 Hz), v is the speed of sound (343 m/s), vo is the speed of the observer (in this case, 0, since the man's velocity and the sound's velocity are in the same direction as he is falling), and vs is the velocity of the source (which is also 0, as the loudspeaker is stationary).

Since we are given f' and f, and we know v, we can solve for vo:

3218.0 = 3000.0 (343 + vo) / 343

3218.0 / 3000.0 = (343 + vo) / 343

1.07267 = (343 + vo) / 343

vo = 1.07267 x 343 - 343

vo = 26.7 m/s (approximately)

Now we apply the equation for free fall to find distance s:

s = (1/2)gt²

To find the time t, we use vo = g × t:

26.7 = 9.8 × t

t ≈ 2.73 s

Plugging the time into the free-fall equation gives us:

s = 0.5 × 9.8 × (2.73)²

s ≈ 36.4 m

So the man has fallen approximately 36.4 meters at the instant he hears the frequency of 3218.0 Hz.