Two strings with circular cross section and made of same material, are stretched to have same amount of tension. A transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section R is $v_1$, and that in the other string having radius of cross section R/2 is $v_2$. Then $\frac{v_2}{v_1}$ =

The equation of a wave travelling on a string is y = sin[20πx + 10πt], where x and t are distance and time in SI units. The minimum distance between two points having the same oscillating speed is :

Two harmonic waves moving in the same direction superimpose to form a wave $x=\mathrm{a} \cos (1.5 \mathrm{t}) \cos (50.5 \mathrm{t})$ where t is in seconds. Find the period with which they beat. (close to nearest integer)

Displacement of a wave is expressed as $x(t)=5 \cos \left(628 t+\frac{\pi}{2}\right) \mathrm{m}$. The wavelength of the wave when its velocity is $300 \mathrm{~m} / \mathrm{s}$ is :

$$(\pi=3.14)$$