Question

The first overtone of a pipe closed at one end resonates with the third harmonic of a string fixed at its ends. The ratio of the speed of sound to the speed of transverse wave travelling on the string is 2 : 1. If the ratio of the length of pipe to the length of string is 1:x. Find x.

Answer 1

Let $v_s$ and $L$ be the speed of sound and length of closed pipe respectively.

Closed organ pipe: Frequency of nth harmonic, ${\nu }_n^{\prime }=\frac{n{v}_s}{4L}$ where $n=2p-1$

For Ist overtone OR third harmonic i.e $n=3$, ${\nu }_3^{\prime }=\frac{3v_s}{4L}$

Transverse wave in string:

Let $v_t$ and $l$ be the velocity of transverse wave and length of the string respectively.

Frequency of mth harmonic, ${\nu }_m=\frac{m{v}_t}{2l}$

For third harmonic i.e $m=3$, ${\nu }_3=\frac{3v_t}{2l}$

Now ${\nu }_3^{\prime }={\nu }_3$

$\frac{3v_s}{4L}=\frac{3v_t}{2l}⟹\frac{v_s}{v_t}=2\frac{L}{l}$

$\frac{2}{1}=2\frac{1}{x}⟹x=1$