Answer to Question #285639 in Mechanics | Relativity for john

Question

A student holds a tuning fork oscillating at 257 Hz. He walks toward a wall at a constant speed of 1.41 m/s.

(a) What beat frequency does he observe between the tuning fork and its echo?

(b) How fast must he walk away from the wall to observe a beat frequency of 4.90 Hz?

33 m/s

Answer 1

v₀ = speed of student

v = speed of sound in air

v = 343 m/s

Let original frequency will be f₀.

Frequency received by wall $f'= f_0(\frac{v}{v-v_0})$

This frequency sound is reflected by the wall and is heard by the student.

Frequency received by student $= f'(\frac{v+v_0}{v})$

$=f_0(\frac{v}{v-v_0})(\frac{v+v_0}{v}) \\ = f_0(\frac{v+v_0}{v-v_0})$

(a) Frequency observed by student $= 257(\frac{343+1.41}{343-1.41}) = 259.12$

Beat frequency = 259.12 -257 = 2.12 Hz

(b) If student walks away from wall, frequency observed by wall $= f_0(\frac{v}{v+v_0}) =f'$

Frequency received after reflection from wall by student $= f'(\frac{v-v_0}{v})$

$=f_0(\frac{v}{v+v_0})(\frac{v-v_0}{v}) \\ = f_0(\frac{v-v_0}{v+v_0})$

Beat frequency = 4.90 Hz

Frequency observed by student will be less than original = 257 – 4.90 = 252.10 Hz

$252.10 = 257(\frac{343-v_0}{343+v_0}) \\ 252.10(343+v_0) = 257(343 -v_0) \\ v_0(252.10+257) = 343(257-252.10) \\ 509.10v_0 = 1680.7 \\ v_0 = 3.3 \;m/s$