Two points are located at a distance of $10\; m$ and $15 \;m$ from the source of oscillation. The period of oscillation is $0.05 \;s$ and the velocity of the wave is $300 \;m/s$. What is the phase difference between the oscillations of the two points?

$A$ wave is propagating along the $x$-axis. The displacement of particles of the medium in the $z$-direction at $t = 0$ is given by: $z = \exp[-(x + 2)^2]$,where $x$ is in meters. At $t = 1 \ s$,the same wave disturbance is given by: $z = \exp[-(2 - x)^2]$. Then,the wave propagation velocity is:

The displacement of a particle in a medium is $y = 10^{-4} \sin(100t + 20x + \frac{\pi}{3}) \ m$,where $t$ is in seconds and $x$ is in metres. The speed of the wave is (in $m/s$)

The distance between two consecutive points with phase difference of $45^{\circ}$ in a wave of frequency $300 \text{ Hz}$ is $4.0 \text{ m}$. The velocity of the travelling wave is (in $\text{km/s}$):

Given below are some functions of $x$ and $t$ to represent the displacement of an elastic wave.
$(i) \, y = 5 \cos (4x) \sin (20t)$
$(ii) \, y = 4 \sin (5x - t/2) + 3 \cos (5x - t/2)$
$(iii) \, y = 10 \cos (252\pi t) \cos (250\pi t)$
$(iv) \, y = 100 \cos (100\pi t + 0.5x)$
State which of these represent:
$(a)$ a travelling wave along $-x$ direction
$(b)$ a stationary wave
$(c)$ beats
$(d)$ a travelling wave along $+x$ direction.
Give reasons for your answers.

The equation of the wave is $Y = 10 \sin \left(\frac{2 \pi t}{30} + \alpha\right)$. If the displacement is $5 \text{ cm}$ at $t = 0$,then the total phase at $t = 7.5 \text{ s}$ will be (Given: $\sin 30^{\circ} = 0.5$)