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Two organ pipes, open at one end but closed at the other, are each 1.03 m long. One is now lengthened by 1.50 cm. Find the beat frequency they produce when playing together in their fundamental.

Sagot :

okpalawalter8

Answer:

[tex]f=1.297Hz[/tex]

Explanation:

From the question we are told that:

Initial pipe length [tex]L_1=1.03m[/tex]

Final pipe length [tex]L_2=1.03+(1.50*10^-2)[/tex]

                             [tex]L_2=1.045m[/tex]

Generally the equation for Frequency f is mathematically given by

 [tex]F=\frac{v}{\lambda}[/tex]

Where

 [tex]\lambda=4L[/tex]

And

V=speed of sound in air

 [tex]v=343m/s[/tex]

Therefore

 [tex]f=\frac{v}{4}{\frac{1}{L_1}-\frac{1}{L_2}}[/tex]

 [tex]f=\frac{343}{4}{\frac{1}{1.03}-\frac{1}{1.045}}[/tex]

 [tex]f=1.297Hz[/tex]

onyebuchinnaji

The beat frequency they produce when playing together in their fundamental is 1.2 Hz.

The given parameters;

  • initial pipe length, L₁ = 1.03 m
  • final pipe length, L₂ = 1.03 m + 1.5 cm = 1.03 m + 0.015 m = 1.045 m

The frequency of wave is calculated as follows;

[tex]v = f \lambda \\\\f = \frac{v}{\lambda }[/tex]

The fundamental frequency of closed pipe is calculated as follows;

[tex]L= \frac{\lambda}{4} \\\\\lambda = 4L\\\\f = \frac{v}{4L}[/tex]

The beat frequency is calculated as follows;

[tex]f_0 - f_1 = \frac{v}{4L_0} - \frac{v}{4L_1} \\\\f_0 - f_1 = \frac{v}{4} (\frac{1}{L_0} - \frac{1}{L_1} )\\\\f_0 - f_1 = \frac{343}{4} (\frac{1}{1.03} - \frac{1}{1.045} )\\\\f_0 - f_1 = 1.2 \ Hz[/tex]

Thus, the beat frequency they produce when playing together in their fundamental is 1.2 Hz.

Learn more here:https://brainly.com/question/23432691

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