Question

An organ pipe $P_{1}$ closed at one end vibrating in its first overtone and another pipe $P_{2}$ open at both ends vibrating in third overtone are in resonance with a given tuning fork. The ratio of the length of $P_{1}$ to that of $P_{2}$ is

• A. $8 / 3$
• B. $3 / 8$
• C. $1 / 2$
• D. $1 / 3$

Answer 1

Answer: (B)

For the organ pipe $P_{1}$ closed at one end the frequency is
$v_{1}=(2 n+1) \frac{v}{4 l_{1}}$
where, variables have their usual meanings.
The first overtone frequency will obtain for $n=1$,
$v_{1}=(2(1)+1) \frac{v}{4 l_{1}} $
$v_{1}=\frac{3 v}{4 l_{1}} \ldots$
For the organ pipe $P_{2}$ open at both ends, the frequency is $v_{2}=\frac{n v}{2 l_{2}}$
where, variables have their usual meanings.
The first overtone frequency will obtain for $n=2$,
hence, the third overtone frequency will obtain for $n=4$,
$v_{2}=\frac{4 v}{2 l_{2}}$
These frequencies are in a resonance with a given tuning fork.
Hence, $v_{1}=v_{2}$
$\frac{3 v}{4 l_{1}}=\frac{4 v}{2 l_{2}}$
$\frac{l_{1}}{l_{2}}=\frac{6}{16}$
$\frac{l_{1}}{l_{2}}=\frac{3}{8}$