Question

The speed of sound in air is $340~ms^{-1}$. The frequency of the fundamental note emitted by a pipe of length 2 m closed at one end is

• A. 42.5 Hz
• B. 85 Hz
• C. 170 Hz
• D. 340 Hz
• E. 21.25 Hz

Hint:

Be careful! If the pipe were open at \textit{both} ends, the formula would be $f = v/2L$, which would give 85 Hz. Always double-check if the pipe is closed or open to avoid a factor-of-two error.

Answer 1

The Correct Option is A

Concept:
Physics - Standing Waves in Organ Pipes.
For a pipe closed at one end, the fundamental frequency ($f_0$) is given by $f_0 = \frac{v}{4L}$.
Step 1: Identify the given values.

• Speed of sound ($v$) = $340~ms^{-1}$
• Length of the pipe ($L$) = 2 m

Step 2: Apply the fundamental frequency formula.
For a pipe closed at one end, the length of the pipe corresponds to a quarter of a wavelength ($\lambda/4$). $$ f_0 = \frac{v}{4L} $$
Step 3: Calculate the frequency.
$$ f_0 = \frac{340}{4 \times 2} $$ $$ f_0 = \frac{340}{8} $$ $$ f_0 = 42.5 \text{ Hz} $$