If the length of a closed organ pipe is 1 m and velocity of sound is 330 m/s, then the frequency for the second note is

4 \times \frac{330}{4}

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3 \times \frac{330}{4}

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2 \times \frac{330}{4}

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2 \times \frac{4}{330}

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Solution

The correct option is B $3 \times \frac{330}{4}$ Hz
For closed pipe $η = \frac{ν}{4 l} = \frac{330}{4} H z$
second note = $3 η_{1} = 3 \times \frac{300}{4}$ Hz

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A closed pipe has length 0.6m. The air inside the pipe is maintained at temperature $27^{\circ} C$ . Calculate the fundamental frequency and the frequency of the next two overtones.
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If the length of a closed organ pipe is 1 m and velocity of sound is 330 m/s, then the frequency for the second note is

The correct option is B 3×3304 HzFor closed pipe η=ν4l=3304Hzsecond note =3η1=3×3004 Hz

The correct option is B $3 \times \frac{330}{4}$ Hz
For closed pipe $η = \frac{ν}{4 l} = \frac{330}{4} H z$
second note = $3 η_{1} = 3 \times \frac{300}{4}$ Hz