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A tube that is open at both ends supports a standing wave with harmonics at 300 Hz and 400 Hz, with no harmonics between. What is the fundamental frequency of this tube?
A. 50 Hz  
B. 100 Hz
C. 150 Hz
D. 200 Hz
E. 300 Hz

Answer :

Answer:

B. 100Hz

Explanation:

The resonant frequencies for a fixed string is given by the formula  nv/(2L).  

Where n is the multiple

v is speed in m/s

The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn

fundamental frequency means n=1

i.e  fn+1 – fn = 400 -300

                      =  100

The fundamental frequency is 100Hz

Frequency modes:

The frequency modes of a wave in a tube with both ends open is given by

[tex]f_n=\frac{nv}{2L}[/tex]

where,

n = 1, 2, 3, 4.............

v is the velocity of the wave

L is the length of the tube

Now, n = 1 gives fundamental frequency

The frequencies for n>1 are called overtones.

The difference between two successive overtones is:

[tex]f_{n+1}-f_n=\frac{v}{2L}=f[/tex]

which means the difference between two successive overtones is equal to the fundamental frequency.

So from the question we get:

[tex]f = 400-300 =100\;Hz[/tex]

is the fundamental frequency

Learn more about fundamental frequency:

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