A. Two out-of-tune flutes play the same note. One produces a tone that has a fre

Question

A. Two out-of-tune flutes play the same note. One produces a tone that has a frequency of 244 Hz, while the other produces 282 Hz. When a tuning fork is sounded together with the 244-Hz tone, a beat frequency of 19.0 Hz is produced. When the same tuning fork is sounded together with the 282-Hz tone, a beat frequency of 19.0 Hz is produced. What is the frequency of the tuning fork?

B. A string that is fixed at both ends has a length of 1.03 m. When the string vibrates at a frequency of 89.6 Hz, a standing wave with five loops is formed. (a) What is the wavelength of the waves that travel on the string? (b) What is the speed of the waves? (c) What is the fundamental frequency of the string?

Explanation / Answer

A)

let,

tuning fork frequency is fc

and

let fa=244 Hz and fb=282 Hz

now,

fbeat=fc-fa=19 Hz

here,

fc-fa=19 Hz or fa-fc=19 Hz

fc=19+fa Hz or fc=fa-19 Hz

fc=19+244 or fc=244-19

fc=263 Hz or fc=225 Hz ------(1)

and

fbeat=fc-fb=19 Hz

here,

fc-fb=19 Hz or fb-fc=19 Hz

fc=19+fb Hz or fc=fb-19 Hz

fc=19+282 or fc=282-19

fc=301 Hz or fc=263 Hz ------(2)

from equation no (1) and (2)

tuning fork frequency fc=263 Hz

B)

length of string L=1.03 m

at n=5loops, frequncy f5=89.6 hz

a)

L=5*lambda/2

wavelength,lambda=2L/5

lambda=2*1.03/5 =0.412 m

b)

speed v=lambda*f5

v=0.412*89.6

v=36.91 m/sec

c)

f5=n*f1

here, fundamental frequency is,f1=f5/n

f1=89.6/5

=17.92 Hz

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