$A$ pipe open at both ends produces a note of frequency $f_1$. When the pipe is kept with $3/4$th of its length in water,it produces a note of frequency $f_2$. The ratio $f_1/f_2$ is

$A$ closed organ pipe has length $L$. The air in it is vibrating in the third overtone with a maximum amplitude $a$. The amplitude at a distance $\frac{L}{7}$ from the closed end of the pipe is:

An open pipe of length $33 \ cm$ resonates with a frequency of $1000 \ Hz$. If the speed of sound is $330 \ m/s$,then this frequency is:

The first overtone frequency of a closed pipe of length $l_{1}$ is equal to the $2^{\text{nd}}$ harmonic frequency of an open pipe of length $l_{2}$. The ratio $\frac{l_{1}}{l_{2}}$ is:

The fundamental frequency of a pipe is $100 \ Hz$ and the next two frequencies are $300 \ Hz$ and $500 \ Hz$. Then:

$A$ 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 $20,000\, Hz$)