If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is: 
1. \(1:2\)
2. \(1:1\)
3. \(\sqrt{2}:1\)
4. \(1:\sqrt{2}\)

A string of length \(l\) is fixed at both ends and is vibrating in second harmonic. The amplitude at antinode is \(2\) mm. The amplitude of a particle at a distance \(l/8\) from the fixed end is:
        
1. \(2\sqrt2~\text{mm}\)
2. \(4~\text{mm}\)
3. \(\sqrt2~\text{mm}\)
4. \(2\sqrt3~\text{mm}\)

If the equation of a wave is represented by:

$y$ $=$ ${10}^{-4}$ $\sin (100t$ $-$ $\frac{x}{10})m$, then the velocity of wave will be:
1.  100 m/s
2.  4 m/s
3.  1000 m/s
4.  0.00 m/s

Two waves have the following equations:

$x_1$ $=$ $a$ $\sin$ $(\omega t$ $+$ ${\varphi }_1)$
$x_2$ $=$ $a$ $\sin$ $(\omega t$ $+$ ${\varphi }_2)$

If in the resultant wave, the frequency and amplitude remain equal to the amplitude of superimposing waves, then the phase difference between them will be:

1.  $\frac{\mathrm{\pi }}{6}$

2. $\frac{2\mathrm{\pi }}{3}$

3. $\frac{\mathrm{\pi }}{4}$

4. $\frac{\mathrm{\pi }}{3}$

If the tension and diameter of a sonometer wire of fundamental frequency n are doubled and density is halved, then its fundamental frequency will become:

1. $\frac{n}{4}$

2.  $\sqrt{2}$ $n$ $$

3.  n 

4.  $\frac{n}{\sqrt{2}}$

If a wave is travelling in a positive X-direction with A = 0.2 m, velocity = 360 m/s, and λ = 60 m, then the correct expression for the wave will be:

The phase difference between two waves, represented by
$\begin{array}{l}{\mathrm{y}}_1={10}^{-6}\sin \{100\mathrm{t}+(\mathrm{x}/50)+0.5\}\mathrm{m}\\ {\mathrm{y}}_2={10}^{-6}\cos \{100\mathrm{t}+\left(\frac{\mathrm{x}}{50}\right)\}\mathrm{m}\end{array}$
where X is expressed in metres and t is expressed in seconds, is approximate:
1. 2.07 radians
2. 0.5 radians
3. 1.5 radians
4. 1.07 radians