The wave described by y=0.25 sin(10$\pi x-2\pi t$), where x and y are in metres and t in seconds, is a wave traveling along the:

1.	-ve x direction with frequency 1 Hz
2.	+ve x direction with frequency π Hz and wavelength λ =0.2 m
3.	+ve x direction with frequency 1 Hz and wavelength λ = 0.2 m
4.	- ve x direction with amplitude 0.25 m and wavelength λ = 0.2 m

 

The velocity of sound in air is:

The speed of sound at a constant temperature depends on:

The displacement of a particle is given by $y=5\times {10}^{-4}\sin (100t-50x)$, where x is in metres and t is in seconds. The velocity of the wave is:

1. 5000 m/sec

2. 2 m/sec

3. 0.5 m/sec

4. 300 m/sec

The wave equations of two particles are given by $y_1=a\sin (\omega t-kx)$, $y_2=a\sin (kx+\omega t)$, then:

The equation of a progressive wave is given by $y=4\sin \left\{\pi \left(\frac{{\displaystyle t}}{{\displaystyle 5}}-\frac{{\displaystyle x}}{{\displaystyle 9}}\right)+\frac{{\displaystyle \pi }}{{\displaystyle 6}}\right\}$.
Which of the following is correct​​​​?
1. v = 5 m / sec
2.  λ = 18 m
3.  a = 0.04 m
4.  n = 50 Hz

Two waves are propagating to the point P along a straight line produced by two sources, A and B, of simple harmonic and equal frequency. The amplitude of every wave at P is ‘a’ and the phase of A is ahead by π/3 than that of B, and the distance AP is greater than BP by 50 cm. If the wavelength is 1 meter, then the resultant amplitude at point P will be: 
1. 2a
2. $a\sqrt{3}$
3. $a\sqrt{2}$
4. a

Tuning fork \(F_1\) has a frequency of 256 Hz and it is observed to produce 6 beats/second with another tuning fork \(F_2\). When \(F_2\) is loaded with wax, it still produces 6 beats/second with \(F_1\). The frequency of \(F_2\) before loading was:
1. 253 Hz
2. 262 Hz
3. 250 Hz
4. 259 Hz

Two tuning forks, A and B, vibrating simultaneously produce 5 beats. The frequency of B is 512 Hz. It is seen that if one arm of A is filed a little, then the number of beats increases. The frequency of A in Hz will be:

The equation of a stationary wave is \(y = 0.8 ~cos\left(\frac{\pi x}{20}\right)sin200(\pi t)\)$$, where \(x\) is in cm and \(t\) is in sec. The separation between consecutive nodes will be:

1. \(20~\text{cm}\)

2. \(10~\text{cm}\)

3. \(40~\text{cm}\)

4. \(30~\text{cm}\)