A tuning fork vibrating at 509 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tuning fork when waves of frequency of 486 Hz reach th…

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asked Feb 12, 2020 234k views

1 Answer

Headwinds · Answer 1 · 2020-02-19T02:46:27+0000

Answer:25.873 m

Step-by-step explanation:

Original frequency
(f_s)=509 Hz

Acceleration due to gravity =9.80 m/s^2

Observed frequency
(f_0) =486 Hz

When the source is receding from stationary observer then

$f_0=f_s\left [ (v)/(v-(-v_s)) \right ]$

$f_0=f_s\left [ (v)/(v+v_s) \right ]$

$v_s=v\left [ (f_s)/(f_0)-1\right ]$

$v_s=340\left [(514)/(483)-1 \right ]$

v_s=21.82 m/s

velocity =(distance)/(tme)

Distance tuning fork has fallen to reach this speed

$\delta y=(v^2-0)/(2a)$

$\delta y=((21.82)^2)/(2* 9.8)$

$\delta y=24.29 m$

Time required to reach the sound to the observer

$t=(\delta y_1)/(v)$

t=(24.29)/(340)

t=0.071 s

During this time fork has traveled an additional distance of

$\delta y_2=v_st+(1)/(2)at^2$

$\delta y_2=21.82* 0.071+(1)/(2)(9.81)(0.071)^2$

$\delta y_2=1.583$

$Total distance=\delta y_1+\delta y_2=24.29+1.583=25.873 m$