The equation of a stationary wave on a string clamped at both ends and vibrating in the third harmonic is $Y = 0.5 \sin(0.314 x) \cos(600 \pi t)$,where $x$ and $y$ are in $cm$ and $t$ is in seconds. The length of the vibrating string is: (in $cm$)

$A$ stretched string is fixed at both ends. It is made to vibrate so that the total number of nodes formed in it is '$x$'. The length of the string in terms of the wavelength of waves formed in it is $(\lambda = \text{wavelength})$

$A$ stretched string is vibrating in its $5^{th}$ harmonic as shown. Consider a particle $1$ (figure). At an instant,this particle is at its mean position and is moving towards its negative extreme. Which of the following sets of particles are in the same phase as particle $1$?

In a vibrating string with fixed ends,the waves are of type:

The pattern of standing waves formed on a stretched string at two instants of time is shown in the figure. The velocity of the two waves superimposing to form stationary waves is $360 \ m/s$ and their frequencies are $256 \ Hz$.
$(a)$ Calculate the time at which the second curve is plotted.
$(b)$ Mark nodes and antinodes on the curve.
$(c)$ Calculate the distance between $A^{\prime}$ and $C^{\prime}$.

If in a stationary wave the amplitude corresponding to an antinode is $4 \,cm$,then the amplitude corresponding to a particle of medium located exactly midway between a node and an antinode is ........... $cm$.