JEE PYQ: Waves And Sound Question 42
Question 42 - 2019 (10 Jan Shift 2)
A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be: (Assume that the highest frequency a person can hear is 20,000 Hz)
(1) 6
(2) 4
(3) 7
(4) 5
Show Answer
Answer: (1)
Solution
If a closed pipe vibration in $N^{\text{th}}$ mode then frequency of vibration $n = \frac{(2N-1)v}{4l} = (2N-1)n_1$
(where $n_1 =$ fundamental frequency of vibration)
Hence $20,000 = (2N - 1) \times 1500$
$\Rightarrow N = 7.1 \approx 7$
$\therefore$ Number of over tones $=$ (No. of mode of vibration) $- 1$
$= 7 - 1 = 6$
BETA
