The fundamental frequency of a closed pipe of length $L$ is equal to the second overtone of a pipe open at both the ends of length $(XL)$. The value of $X$ is (Neglect end correction).

$A$ closed organ pipe has length $l$. The air in it is vibrating in the $3^{rd}$ overtone with a maximum displacement amplitude $a$. The displacement amplitude at a distance $l/7$ from the closed end of the pipe is:

One insulated conductor from a household extension cord has a mass per unit length of $\mu$. $A$ section of this conductor is held under tension between two clamps. $A$ subsection is located in a magnetic field of magnitude $B$ directed perpendicular to the length of the cord. When the cord carries an $AC$ current of $i$ at a frequency of $f$,it vibrates in resonance in its simplest standing-wave vibration state. Determine the relationship that must be satisfied between the separation $d$ of the clamps and the tension $T$ in the cord.

In the fundamental mode,the time required for a sound wave to reach the closed end of a pipe filled with air is $t$ seconds. The frequency of vibration of the air column is:

$A$ pipe open at both ends and a pipe closed at one end have the same length. The ratio of the frequencies of the air columns in their $p^{\text{th}}$ overtone respectively is

$A$ resonance air column of length $20 \ cm$ resonates with a tuning fork of frequency $250 \ Hz$. The speed of sound in air is .... $m/s$.