$A$ wave is represented by the equation $y = 10 \sin 2\pi(100t - 0.02x) + 10 \sin 2\pi(100t + 0.02x)$. The maximum amplitude and loop length are respectively:

$A$ standing wave pattern of amplitude $A$ in a string of length $L$ shows $2$ nodes (plus those at two ends). If one end of the string corresponds to the origin and $v$ is the speed of the progressive wave,the disturbance in the string could be represented (with appropriate phase) as:

$A$ standing wave is represented by $Y = A \sin(100t) \cos(0.01x)$,where $Y$ and $A$ are in millimetres,$t$ is in seconds,and $x$ is in metres. The velocity of the wave is ..... $m/s$.

The transverse displacement of a string of a linear density $0.01 \text{ kg m}^{-1}$,clamped at its ends is given by $Y_{(x, t)} = 0.03 \sin \left(\frac{2 \pi x}{3}\right) \cos (60 \pi t)$,where $x$ and $y$ are in metres and time $t$ is in seconds. The tension in the string is: (in $\text{ N}$)

$A$ string fixed at both ends vibrates in $5$ loops as shown in the figure. The total number of nodes and antinodes respectively are

Two travelling waves of equal amplitudes and equal frequencies move in opposite directions along a string. They interfere to produce a stationary wave whose equation is given by $y = (10 \cos \pi x \sin \frac{2 \pi t}{T}) \, cm$. The amplitude of the particle at $x = \frac{4}{3} \, cm$ will be ........ $cm$.