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

For a transverse wave travelling along a straight line,the distance between two peaks (crests) is $5 \, m$,while the distance between one crest and one trough is $1.5 \, m$. The possible wavelengths (in $m$) of the waves are

$A$ simple wave motion is represented by $y=5(\sin 4 \pi t+\sqrt{3} \cos 4 \pi t)$. Its amplitude is

$A$ person is producing a wave in a string by moving his hand first up and then down. If the frequency is $\frac{1}{8} \, Hz$,then find out the time taken by a particle which is at a distance of $9 \, m$ from the source to move to the lower extreme for the first time in seconds. (Given $\lambda = 24 \, m$)

The equations of two simple harmonic waves are given by $Y_1 = 2 \sin 8 \pi \left(\frac{t}{0.2} - \frac{x}{2}\right) \text{ m}$ and $Y_2 = 4 \sin 8 \pi \left(\frac{t}{0.16} - \frac{x}{1.6}\right) \text{ m}$. Then both waves have:

The equation of a travelling wave is given by $y = 0.5 \sin(20x - 400t)$,where $x$ and $y$ are in meters and $t$ is in seconds. The velocity of the wave is .... $m/s$.