For a stationary wave represented by the equation $y = 0.15 \sin(5x) \cos(300t) \, \text{m}$,what is the wavelength of the wave (in $, \text{m}$)?

Two progressive waves $Y_1 = \sin 2 \pi \left( \frac{t}{0.4} - \frac{x}{4} \right)$ and $Y_2 = \sin 2 \pi \left( \frac{t}{0.4} + \frac{x}{4} \right)$ superpose to form a standing wave. $x$ and $y$ are in $SI$ units. The amplitude of the particle at $x = 0.5 \ m$ is $\left[ \sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}} \right]$.

The stationary wave $y = 2a \sin kx \cos \omega t$ in a stretched string is the result of the superposition of $y_1 = a \sin(kx - \omega t)$ and

For a stationary wave, $y = 12 \cos \left( \frac{\pi x}{10} \right) \sin(36\pi t) \text{ cm}$, the distance between a node and the successive antinode is (in $\text{ cm}$)

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 $cm$)

In stationary waves,