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Q.
Two open organ pipes of length $25\, cm$ and $25.5\, cm$ produce $0.1$ beat/sec. The velocity of sound will be :
Punjab PMETPunjab PMET 2004Electromagnetic Waves
Solution:
Here : Length of first pipe $l_{1}=25\, cm$
Length of second pipe $l_{2}=25.5\, cm$
Frequency $f$ beat $=0.1$ beat/sec
Frequence of first pipe $f_{1}$ is given by,
$\frac{v}{2 l_{1}}=\frac{v}{2 \times 25}=\frac{v}{50} Hz$
Frequency of second pipe $f_{2}$ is given by,
$\frac{v}{2 l_{2}}=\frac{v}{2 \times 25.5}=\frac{v}{51} Hz$
(where $v$ is velocity of sound)
Hence, the frequency
$f =f_{1}-f_{2}$
$0.1 =\frac{v}{50}-\frac{v}{51}=\frac{v}{50 \times 51}$
or $v =0.1 \times 50 \times 51$
$v =255\, m / s$