Waves Test - 58...

The equation of wave is given by $$y = 10 \sin(2\pi t/45 + \alpha)$$. If the displacement is$$ 5 cm$$ at $$=0$$, then the total phase at $$t = 7.5 s$$ will be:

A $$4kg$$ particle is moving along the x-axis under the action of the force which varies with distance $$x$$ given by $$F=-\left( \dfrac { { \pi  }^{ 2 } }{ 16 }  \right) x$$ Newton. At $$t=2sec$$, the particle passes through the origin and at $$t=10sec$$, its speed is $$ 4\sqrt { 2 } m/s $$. The amplitude of the motion is

What is the period of a vibrating particle executing S.H.M.. Which has acceleration $$ 0.12 m/s^2 $$ when its displacement is $$ 3 \times 10^{-2} m? $$

Two waves represented as 
$$y_1 = a \sin (\omega t + {\pi}/6)$$, $$y_2 = a  \cos  \omega t$$
The resultant amplitude is-

Two particles in the path of a wave, of velocity $$360\ m/s$$ and frequency $$2000\ Hz$$, differ in phase by $$120^{o}$$. The distance between the particles is

A particle executes simple harmonic motion and is located x = a, b and c at time$$ \ t_0 \ 2t_0 \ and \ 3t_0 $$ respectively.The frequency of the oscillation is:- 

On the superposition of the two waves given as $${ y }_{ 1 }={ A }_{ 0 }sin\left( \omega t-kx \right) $$ and $${ y }_{ 2 }={ A }_{ 0 }cos\left( \omega t-kx+\dfrac { \pi  }{ 6 }  \right) $$. the resultant amplitude of oscillations will be 

$$If x=a\sin { (\omega +\frac { \pi  }{  6}  })$$  and $$x^2=a\cos \omega { t }$$ ,then what is the phase difference between the two waves:

Two waves are represented by the equations 
$$y_1 = a sin (\omega  + kx + 0.57 )m$$ and
$$y_2 = a cos (\omega t + kx)m$$
where x is in meter and t in sec. The phase difference between them is 

Two particles undergo SHM along the same line with the same time period (TT They will cross each other after a further time