Tag: oscillation and waves

A wave has a wavelength of 3m. The distance between a crest and adjacent trough is

A source oscillates with a frequency 25 Hz and the wave propagates with 300 m/s. Two points A and B are located at distances 10 m and 16 m away from the source. The phase difference between A and B is 

Two simple harmonic motions are represented by the equations 
$y _1=10\sin \left(3\pi t+\dfrac{\pi}{4}\right)$
and $y _2=5(3\sin 3\pi t+\sqrt 3 \cos 3\pi t)$ Their amplitudes are in the ratio of :

For the travelling harmonic wave  $y(x,t)=2.0 cos $ $ 2\pi $ (10t-0.0080 x+0.35 ) where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of $x$

Vibrations of period 0.25 s propagate along a straight line at a velocity of 48 cm/s. One second after the emergence of vibrations at the initial point, displacement of the point, 47 cm from it is found to be 3 cm. Then,

A wave travelling in positive X-direction with A = 0.2 m velocity = 360 m/s and $\lambda$= 60 m, then correct expression for the wave is : -

If two waves, each of intensity ${I} _{0}$, having the same frequency but differing by a constant phase angle of ${60}^{o}$, superpose at a certain point in space, then the intensity of resultant wave is:

The displacement of an elastic wave is given by the function $y= 3\ sin \omega t+4\ cos\omega t$, where $y$ is in $cm$ and $t$ is in $s$. The resultant amplitude is 

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 distance $l/8$ from the fixed end is :

Equations of a stationary wave and a travelling wave are $y _1=1\,sin(kx)\,cos (\omega t)$ and $y _2=a\,sin\,(\omega t-kx)$.The phase difference between two points $x _1=\dfrac{\pi}{3k}$ and $x _2=\dfrac{3 \pi}{2k}$ is $\phi _1$ for the first wave and $\phi _2$ for the second wave.The ratio $\dfrac{\phi _1}{\phi _2}$ is