Respuesta :
Explanation:
It is given that,
Fundamental frequency, f = 220 Hz
(a) We know that at 0 degrees, the speed of sound in air is 331 m/s.
For open pipe, [tex]\lambda=2l[/tex]
l is the length of pipe
Also,
[tex]v=f\lambda[/tex]
[tex]l=\dfrac{v}{2f}=\dfrac{331}{2\times 220}=0.75\ m[/tex]
(b) Let f' is the fundamental frequency of the pipe at 30 degrees and v' is its speed.
[tex]v'=331\sqrt{1+\dfrac{T}{273}}[/tex]
[tex]v'=331\sqrt{1+\dfrac{30}{273}}[/tex]
v' = 348.71 m/s
So, [tex]f'=\dfrac{v'}{\lambda}[/tex]
[tex]f'=\dfrac{348.71}{2\times 0.75}[/tex]
f' = 232.4 Hz
Hence, this is the required solution.
Answer:284.4 Hz
Explanation:
At [tex]o^{\circ}C[/tex] speed of sound is 331 m/s
and we know
[tex]Velocity\left ( v\right )=frequency\left ( f\right )\times wavelength\left ( \lambda\right )[/tex]
[tex]331=220\left ( 2L\right )[/tex]------[tex]\left ( \lambda =2L\right )[/tex]
L=0.613 m
[tex]\left ( b\right )[/tex]
[tex]v=v_0\sqrt{1+\frac{T}{273}}[/tex]
where velocity of sound at[tex]t=0^{\circ}C[/tex]
[tex]v=331\sqrt{1+\frac{30}{273}}[/tex]
v=348.7 m/s
for frequency
[tex]v=f\lambda [/tex]
[tex]348.7=f\times 2\times 0.613[/tex]
f=284.4 Hz
