Two closed pipes when sounded simultaneously in their fundamental modes produce 6 beats per second. If the length of the shorter pipe is 150 cm, then the length of the longer pipe is: \[ \text{(Speed of sound in air = 336 ms}^{-1}) \]
Two closed pipes when sounded simultaneously in their fundamental modes produce 6 beats per second. If the length of the shorter pipe is 150 cm, then the length of the longer pipe is: \[ \text{(Speed of sound in air = 336 ms}^{-1}) \]
Show Hint
- 168 cm
- 184 cm
- 176 cm
192 cm
The Correct Option is A
Solution and Explanation
Step 1: Frequency of Fundamental Mode for a Closed Pipe For a closed pipe, the fundamental frequency is given by: \[ f = \frac{v}{4L} \] where: - \( v = 336 \) m/s (speed of sound in air), - \( L \) is the length of the pipe.
Step 2: Compute Frequency for the Shorter Pipe For the shorter pipe of length 150 cm (\( L_1 = 1.5 \) m): \[ f_1 = \frac{v}{4L_1} = \frac{336}{4 \times 1.5} = \frac{336}{6} = 56 \text{ Hz} \]
Step 3: Determine Frequency of Longer Pipe Given the beat frequency is 6 Hz, the frequency of the longer pipe is: \[ f_2 = f_1 + 6 = 56 + 6 = 62 \text{ Hz} \]
Step 4: Compute Length of the Longer Pipe Using the formula for the fundamental frequency: \[ L_2 = \frac{v}{4f_2} = \frac{336}{4 \times 62} = \frac{336}{248} = 1.68 \text{ m} = 168 \text{ cm} \] Thus, the correct answer is: \[ \mathbf{168 \text{ cm}} \]
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