1 year ago

Caillou wait no what

Physics

2 answers:

Tomtit [17]1 year ago

5 0

Answer:ayo

Explanation:

zlopas [31]1 year ago

3 0

Ayo what the heck is this?

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Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A pipe closed at both ends can have standing wav

uranmaximum [27]

Answer:

68.8 Hz

137.6 Hz, 206.4 Hz

Explanation:

L = Length of tube = 2.5 m

v = Velocity of sound in air = 344 m/s

Distance between nodes is given by

$L=\dfrac{\lambda}{2}+m\dfrac{\lambda}{2}\\\Rightarrow \dfrac{\lambda(n+1)}{2}=L\\\Rightarrow \lambda=\dfrac{2L}{n+1}$

Where n = 0, 1, 2, 3, ...

Making n+1 = n

$\lambda=\dfrac{2L}{n}$

where n = 1, 2, 3 .....

For fundamental frequency n = 1

$\lambda=\dfrac{2\times 2.5}{1}\\\Rightarrow \lambda=5\ m$

Frequency is given by

$f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{344}{5}\\\Rightarrow f=68.8\ Hz$

The fundamental frequency is 68.8 Hz

First overtone

$2f=2\times 68.8=137.6\ Hz$

Second overtone

$3f=3\times 68.8=206.4\ Hz$

The overtones are 137.6 Hz, 206.4 Hz

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