Two strings $(A, B)$ having linear densities $\mu_A=2 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ and, $\mu_B=4 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ and lengths $L_A=2.5 \mathrm{~m}$ and $L_B=1.5 \mathrm{~m}$ respectively are joined. Free ends of $A$ and $B$ are tied to two rigid supports $C$ and $D$, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from $C$ and $D$ ends, take time $t_1$ and $t_2$, respectively, to reach the joint. The ratio $t_1 / t_2$ is:

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, $y_1(x, t) = 4 \sin (kx - \omega t)$ and $y_2(x, t) = 2 \sin (kx - \omega t + \frac{2\pi}{3})$, are:

(Take the angular frequency of initial waves same as $\omega$)

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 :