A string wave equation is given $y=0.002 \sin (300 t-15 x)$ and mass density is $\mu=\frac{0.1\, kg }{m}$. Then find the tension in the string, (in $N$)

If ${n_1},{n_2},{n_3}.........$ are the frequencies of segments of a stretched string, the frequency $n$ of the string is given by

A string is stretched between two fixed points separated by $75\,cm$ . It is observed to have resonant frequencies of $420\,Hz$ and $315\,Hz$ . There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is  ..... $Hz$

The length of the wire shown in figure between the pulleys is $1.5\, m$ and its mass is  $12.0\,g$. The frequency of vibration with which the wire vibrates in three loops forming antinode at the mid point of the wire is $(g = 9.8 \,m/s^2)$

Two uniform strings of mass per unit length $\mu$ and $4 \mu$, and length $L$ and $2 L$, respectively, are joined at point $O$, and tied at two fixed ends $P$ and $Q$, as shown in the figure. The strings are under a uniform tension $T$. If we define the frequency $v_0=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$, which of the following statement($s$) is(are) correct?

$(A)$ With a node at $O$, the minimum frequency of vibration of the composite string is $v_0$

$(B)$ With an antinode at $O$, the minimum frequency of vibration of the composite string is $2 v_0$

$(C)$ When the composite string vibrates at the minimum frequency with a node at $O$, it has $6$ nodes, including the end nodes

$(D)$ No vibrational mode with an antinode at $O$ is possible for the composite string

Two similar sonometer wires given fundamental frequencies of $500Hz$. These have same tensions. By what amount the tension be increased in one wire so that the two wires produce $5$ beats/sec .... $\%$