Wave motion

This post only for maharastra HSC students. Other may find this stuff usefull
Wave motion
Solution of all problems:

Q 1)  The equation of simple harmonic progressive wave is given by

$ y=4\times \sin\pi\times \left( \dfrac {t}{0.02}-\dfrac {x}{75}\right) $   cm.

Find the Displacement and velocity of the particle at Distance of 50 cm from the origin and at the instant 0.1 second.

Given Data: 

Distance of particle from Origin = x =50 cm

time = t = 0.1

To Find Out: 1) Displacement of particle = y=?

                      2) velocity of particle = Vp = ?

Formula:

       $ 1. y=A\sin 2\times \pi \times \left( \dfrac {t}{T}-\dfrac {x}{A}\right)  .......................(1) $

       $ 2.V=\omega \sqrt {A^{2}-y^{2}}       ...........................................................................(2)   $

Solution:

let p be the particle at a distance x from origin oscillating along y axis.

$\therefore $ Displacement of particle at an instant t = 0.1 sec is given by

$y=4\times \sin\pi\times \left( \dfrac {t}{0.02}-\dfrac {x}{75}\right)     ....(Given )   .........(3) $

( In below fig, x denote the distance of particle from origin and y indicate the displacement of the particle along y axis path. blue Dotted line indicate the path of vibrating particle P) 

put x= 50 & t = 0.1 sec in eq (3)

$\therefore $ equation (3) becomes

$ y=4\sin \pi \left( \dfrac {0.1}{0.02}-\dfrac {50}{75}\right) $

$   \therefore  y=4\sin \pi \left( 5-\dfrac {2}{3}\right) $

$\therefore y=4\times \sin \left[ \dfrac {13\times \pi }{3}\right] $

$ \therefore y=4\sin \left( 4\pi +\dfrac {\pi }{3}\right) $

$\because \sin \left( 2n\pi +\theta \right) =\sin \theta  $

$ \therefore  y=4\sin \left( \dfrac {\pi }{3}\right) $

 $\therefore y=4\dfrac {\sqrt {3}}{2}\\$

$ \therefore y=2\sqrt {3}\\$

$ \therefore y=2\times 1.732 $

$ \therefore  y=3.464  cm    .......................(4)  $

now given equation of displacement

$ y=4\times \sin\pi\times \left( \dfrac {t}{0.02}-\dfrac {x}{75}\right) $  

$ \therefore  y=4\times \sin\2pi\times \left( \dfrac {t}{0.04}-\dfrac {x}{150}\right) $  

comparing above equation with standard equation i.e equation (1)

$\therefore$ we get

$A = 4 ,T = 0.04 $

Now velocity of particle can be given by

  $ V=\omega \sqrt {A^{2}-y^{2}}$

$ \therefore V=\dfrac {2\pi }{T}\left( \sqrt {A^{2}-y^{2}}\right) $               . $ \because w=\dfrac {2\pi }{T} $ 

$ \therefore V=\dfrac {2\times 3.142}{0.04}\sqrt {4^{2}-\left( 3\cdot 464\right) ^{2}}$

$\therefore V= 314 cm/sec $

$\therefore V=3.14 m/sec $

Q2.The frequency of a tuning fork is 256 Hz and velocity of sound in air is 350 m/s. Find the Distance covered by the wave when the fork complete 16 vibrations.

Given Data:
 frequency of a tuning fork = $ n = 256 Hz $
 Velocity of Sound $= V= 350 m/s $
 Number of vibration $ N = 16 $
To find out :
1) Distance covered by the wave = X =?

Formula: 1 ) V = n  $\lambda$
                2)   X = $ N \lambda$
Solution:
First of all we need to find wavelength of the wave.
$\therefore V=n\lambda$
$\therefore350=256$x$\lambda$

$\therefore \lambda =\dfrac {350}{256}$
$\therefore \lambda=1.3671875$
Now Distance covered by the wave is
X=$ N \lambda$
\Therefore X$=20 X 1.3671875
Other solution  will update soon.
thanks for your patience.