Question

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

Answer 1

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