Question #285639

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


1
Expert's answer
2022-01-10T09:08:46-0500

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=f0(vvv0)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(v+v0v)= f'(\frac{v+v_0}{v})

=f0(vvv0)(v+v0v)=f0(v+v0vv0)=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(343+1.413431.41)=259.12= 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 =f0(vv+v0)=f= f_0(\frac{v}{v+v_0}) =f'

Frequency received after reflection from wall by student =f(vv0v)= f'(\frac{v-v_0}{v})

=f0(vv+v0)(vv0v)=f0(vv0v+v0)=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(343v0343+v0)252.10(343+v0)=257(343v0)v0(252.10+257)=343(257252.10)509.10v0=1680.7v0=3.3  m/s252.10 = 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


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