A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin
g fork when waves of frequency of 490 Hz reach the release point? (Take the speed of sound in air to be 343 m/s.)
1 answer:
Answer:
Explanation:
given,
tuning fork vibration = 508 Hz
accelerates = 9.80 m/s²
speed of sound = 343 m/s
observed frequency = 490 Hz


![v_s = v[\dfrac{f_s}{f_o}-1]](https://tex.z-dn.net/?f=v_s%20%3D%20v%5B%5Cdfrac%7Bf_s%7D%7Bf_o%7D-1%5D)
![= 343[\dfrac{508}{490}-1]](https://tex.z-dn.net/?f=%3D%20343%5B%5Cdfrac%7B508%7D%7B490%7D-1%5D)

distance the tunning fork has fallen


=8.1 m
now, time required for the observed will be

now, for the distance calculation


=0.293 m
total distance
= 8.1 + 0.293 = 8.392 m
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