$A$ tuning fork gives $5$ beats per second with $40 \ cm$ length of a sonometer wire. If the length of the wire is shortened by $1 \ cm$,the number of beats is still the same. The frequency of the fork is (in $Hz$)

Two uniform stretched strings $A$ and $B$, made of steel, are vibrating under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the lengths of the strings is

$A$ horizontal stretched string,fixed at two ends,is vibrating in its fifth harmonic according to the equation,$y(x, t) = (0.01 \ m) \sin[(62.8 \ m^{-1}) x] \cos[(628 \ s^{-1}) t]$. Assuming $\pi = 3.14$,the correct statement$(s)$ is (are) :
$(A)$ The number of nodes is $5$.
$(B)$ The length of the string is $0.25 \ m$.
$(C)$ The maximum displacement of the midpoint of the string from its equilibrium position is $0.01 \ m$.
$(D)$ The fundamental frequency is $100 \ Hz$.

When tension $T$ is applied to a sonometer wire of length $\ell$,it vibrates with the fundamental frequency $n$. Keeping the experimental setup same,when the tension is increased by $8 \ N$,the fundamental frequency becomes three times the earlier fundamental frequency $(n)$. The initial tension applied to the wire in newton was:

$A$ steel rod $100 \ cm$ long is clamped at its mid-point. The fundamental frequency of longitudinal vibrations of the rod is given to be $2.53 \ kHz$. What is the speed of sound in steel in $km/s$ (in $.06$)?

The fundamental frequency of a sonometer wire is $n$. If the tension is increased $3$ times,the length is increased $3$ times,and the diameter is increased $2$ times,what will be the new frequency?