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: \(y=10^{-4}~ \mathrm{sin}\left(100t-\dfrac{x}{10}\right)~\text m,\) where \(x \) is in meters and \(t\) in seconds, then the velocity of the wave will be:

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.  $\frac{n}{\sqrt{2}}$

The phase difference between two waves, represented by
\(y_1= 10^{-6}\sin \left\{100t+\left(\frac{x}{50}\right) +0.5\right\}~\text{m}\)
\(y_2= 10^{-6}\cos \left\{100t+\left(\frac{x}{50}\right) \right\}~\text{m}\)
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