$A$ piano wire with a diameter of $0.90 \ mm$ is replaced by another wire of diameter $0.93 \ mm$ of the same material. If the tension of the wire is kept the same,then the percentage change in the frequency of the fundamental tone is

$L$ length and $M$ mass uniform wire is stretched between two fixed points with tension $F$. $A$ sound of frequency $\mu$ is impressed on it. The maximum vibrational energy exists in the wire when $\mu$ =

If the length of a stretched string is shortened by $x \%$ and the tension is increased by $44 \%$,then the ratio of the initial and final fundamental frequencies is $1:2$. The value of '$x$' is:

$A$ string is clamped at both ends and it is vibrating in its $4^{th}$ harmonic. The equation of the stationary wave is $Y = 0.3 \sin(0.157 x) \cos(200\pi t)$. The length of the string is ..... $m$ (all quantities are in $SI$ units).

Two uniform wires of same material are vibrating under the same tension. If the first overtone of the first wire is equal to the second overtone of the second wire and the radius of the first wire is twice the radius of the second wire, then the ratio of the length of the first wire to the second wire is:

$A$ tuning fork gives $4$ beats with $50 \, cm$ length of a sonometer wire. If the length of the wire is shortened by $1 \, cm$,the number of beats remains the same. The frequency of the fork is ............. $Hz$.