A tube of length $1.05\text{ m}$ is closed at one end. If the velocity of sound in air is $336\text{…
A tube of length $1.05\text{ m}$ is closed at one end. If the velocity of sound in air is $336\text{ ms}^{-1}$, then the fundamental and the next higher overtone (in Hz) are respectively
80, 160
80, 240
160, 320
160, 480
Solution
Fundamental frequency, $f_1 = \frac{v}{4l} = \frac{336}{4 \times 1.05} = 80\text{ Hz}$
Next higher overtone will be at frequency,
$f_3 = 3f_1 = 240\text{ Hz}$