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3 years ago

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A 0.25-m string, vibrating in its sixth harmonic, excites a 0.96-m pipe that is open at both ends into its second overtone reson

ance. The speed of sound in air is 345 m/s. The common resonant frequency of the string and the pipe is closest to
Answer please with procedure

a. 630 Hz.
b. 450 Hz.
c. 700 Hz.
d. 540 Hz.
e. 360 Hz.

1 answer:

Andrews [41]3 years ago

7 0

Answer:

option D

Explanation:

given,

length of the pipe, L = 0.96 m

Speed of sound,v = 345 m/s

Resonating frequency when both the end is open

$f = \dfrac{nv}{2L}$

n is the Harmonic number

2nd overtone = 3rd harmonic

so, here n = 3

now,

$f = \dfrac{3\times 345}{2\times 0.96}$

f = 540 Hz

The common resonant frequency of the string and the pipe is closest to 540 Hz.

the correct answer is option D

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