Waves And Sound Question 5
Question 5 - 2024 (31 Jan Shift 1)
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $60 \mathrm{~cm}$, the length of the closed pipe will be :
(1) $60 \mathrm{~cm}$
(2) $45 \mathrm{~cm}$
(3) $30 \mathrm{~cm}$
(4) $15 \mathrm{~cm}$
Show Answer
Answer: (4)
Solution:
$\frac{\lambda}{4}=\mathrm{L}_{1} \quad 2\left(\frac{\lambda}{2}\right)=\lambda$
$\mathrm{v}=\mathrm{f} \lambda \quad \mathrm{f}{2}=\frac{2 \mathrm{v}}{2 \mathrm{~L}{2}}$
$\mathrm{v}=\mathrm{f}{1}\left(4 \mathrm{~L}{1}\right) \quad \mathrm{f}{2}=\frac{\mathrm{v}}{\mathrm{L}{2}}$
$\mathrm{f}{1}=\frac{\mathrm{v}}{4 \mathrm{~L}{1}}$
$\mathrm{f}{1}=\mathrm{f}{2}$
$\frac{\mathrm{v}}{4 \mathrm{~L}{1}}=\frac{\mathrm{v}}{\mathrm{L}{2}}$
$\Rightarrow \mathrm{L}{2}=4 \mathrm{~L}{1}$
$60=4 \times \mathrm{L}_{1}$
$\mathrm{L}_{1}=15 \mathrm{~cm}$
