The equation of a stationary wave is given by $y = 0.8 \cos \left( \frac{\pi x}{20} \right) \sin (200 \pi t) \text{ cm}$. What is the distance between two consecutive nodes in cm?

Two progressive waves are travelling towards each other with velocity $50 \,m/s$ and frequency $200 \,Hz$. The distance between two consecutive antinodes is (in $\,m$)

$A$ $20 \ cm$ long string,having a mass of $1.0 \ g$,is fixed at both the ends. The tension in the string is $0.5 \ N$. The string is set into vibrations using an external vibrator of frequency $100 \ Hz$. Find the separation (in $cm$) between the successive nodes on the string.

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 $\left[ \sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}} \right]$.

$A$ wave $y = a \sin(\omega t - kx)$ on a string meets with another wave producing a node at $x = 0$. Then the equation of the unknown wave is

$A$ string is rigidly tied at two ends and its equation of vibration is given by $y = \sin(2\pi x) \cos(2\pi t)$. Then the minimum length of the string is .... $m$