If the length of stretched string is shortened by $40\%$ and the tension is increased by $44\%$, then the ratio of the final and initial fundamental frequencies is

A student is performing an experiment using a resonance column and a tuning fork of frequency $244 s ^{-1}$. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is $(0.350 \pm 0.005) m$, the gas in the tube is

(Useful information) : $\sqrt{167 R T}=640 j^{1 / 2} mole ^{-1 / 2} ; \sqrt{140 RT }=590 j ^{1 / 2} mole ^{-1 / 2}$. The molar masses $M$ in grams are given in the options. Take the value of $\sqrt{\frac{10}{ M }}$ for each gas as given there.)

The equation of a wave on a string oflinear mass density $0.04$ $kgm^{-1}$ is given by

$y = 0.02sin\left[ {2\pi \left( {\frac{t}{{0.04\left( s \right)}} - \frac{x}{{0.50\left( m \right)}}} \right)} \right]m$ The tension in the string is  .... $N$

To increase the frequency from $100 Hz$ to $400 Hz$ the tension in the string has to be changed by ..... $times$

A vibrating string of certain length $l$ under a tension $T$ reasonates with a mode corresponding to the first overtone (third harmonic) of an air column of length $75$ $cm$ inside a tube closed at one end. The string also generates $4$ beats per second when excited along with a tuning fork of frequency $n$. Now when the tension of the string is slightly increased the number of beats reduces to $2$ per second. Assuming the velocity of sound in air to be $340$ $m/s$, the frequency $n$ of the tuning fork in $Hz $ is

Standing waves are produced in $10 \,m$ long stretched string fixed at both ends. If the string vibrates in $5$ segments and wave velocity is $20 \,m / s$, the frequency is ....... $Hz$