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nydimaria [60]
4 years ago
9

A woman drops a vibrating tuning fork, which is vibrating at 513 Hertz, from a tall building. Through what distance (in m) has t

he tuning fork fallen when the frequency detected at the starting point is 489 Hertz? (Assume the speed of sound in air is 343 m/s.)
Physics
1 answer:
umka21 [38]4 years ago
6 0

Answer:

h = 15.34 m

Explanation:

given,

tuning fork vibration = 513 Hz

speed of sound = 343 m/s

frequency after deflection = 489 Hz

the source (the fork) moves away from the observer, its speed increases and hence the apparent frequency decreases

f_{apparent} = \dfrac{v}{v+u}f_0

489 = \dfrac{343}{343+u}\times 513

0.953 = \dfrac{343}{343+u}

343+u = \dfrac{343}{0.953}

343+u = 359.92

u = 16.92 m/s

height of the building

v² = u² + 2 g s

16.92² = 2 x 9.8 x h

h = 14.61 m

time taken by sound to reach observer

t = \dfrac{14.61}{343}

t =0.0426\ s

in this time tuning fork has fallen one more now,

h' = u t + \dfrac{1}{2}gt^2

h' = 16.92\times 0.0426 + \dfrac{1}{2}\times 9.8 \times 0.0426^2

h' = 0.7296 m = 0.73 m

total distance

      h = 14.61 + 0.73

      h = 15.34 m

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