The amplitude of a wave,represented by the displacement equation $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$,will be:

The equations of two waves are given by
$y_{1}=5 \sin 2 \pi(x-v t) \, cm$
$y_{2}=3 \sin 2 \pi(x-v t+1.5) \, cm$
These waves are simultaneously passing through a string. The amplitude of the resulting wave is.........$cm$.

If $y_1 = 5 \text{ mm} \sin(\pi t)$ is the equation of oscillation of source $S_1$ and $y_2 = 5 \text{ mm} \sin(\pi t + \pi/6)$ is that of $S_2$,and it takes $1 \text{ s}$ and $0.5 \text{ s}$ for the transverse waves to reach point $A$ from sources $S_1$ and $S_2$ respectively,then the resulting amplitude at point $A$ is .... $\text{mm}$.

Two waves represented by $y_1 = 10 \sin(200\pi t)$ and $y_2 = 20 \sin(200\pi t + \pi/2)$ are superimposed at any point at a particular instant. The amplitude of the resultant wave is:

Three waves of equal frequency having amplitudes $10 \,\mu m, 4 \,\mu m$ and $7 \,\mu m$ arrive at a given point with successive phase difference of $\frac{\pi}{2}$. The amplitude of the resulting wave in $\mu m$ is given by

Sound waves are passing through two routes—one in a straight path and the other along a semicircular path of radius $r$—and are again combined into one pipe and superposed as shown in the figure. If the velocity of sound waves in the pipe is $v$, then frequencies of resultant waves of maximum amplitude will be integral multiples of