Home
Class 11
PHYSICS
Two waves are expressed as, y(1) = a...

Two waves are expressed as,
`y_(1) = a sin omega_(1) ((x)/(c)-t) and y_(2) = a sin omega_(2) ((x)/(c)-t)`
Find the resultant displacement due to superposi-tion of the two waves .

Text Solution

Verified by Experts

Two waves are expressed as,
` y_(1) = asin omega_(1) ((x)/(c)-t) and y_(2) = a sin omega_(2) ((x)/(c)-t)`
So, resultant displacement due to superposition of these two waves is,
` y = y_(1) + y_(2) = asin omega_(1) ((x)/(c)-t) + asin omega_(2) ((x)/(c) -t)`
` = 2 a sin[((omega_(1) + omega_(2))(x)/(c)-(omega_(1)+omega_(2))t)/(2)] xxcos[((omega_(1)-omega_(2))(x)/(c)-(omega_(1)-omega_(2))t)/(2)]`
`= 2 a sin[((omega_(1)+omega_(2)))/(2c)-((omega_(1)+omega_(2))t)/(2)]xxcos[((omega_(1)-omega_(2))x)/(2c)-((omega_(1)-omega_(2))t)/(2)]`
So, `y = A sin [((omega_(1) + omega_(2))x)/(2c) - ((omega_(1) + omega_(2))t)/(2)]`
Where, `a = 2 a cos [((omega _(1)-omega_(2))x)/(2c) - ((omega_(1) - omega_(2))t)/(2)]`
Promotional Banner

Topper's Solved these Questions

  • SUPERPOSITION OF WAVES

    CHHAYA PUBLICATION|Exercise Examination archive with solutions (WBJEE)|5 Videos

  • SUPERPOSITION OF WAVES

    CHHAYA PUBLICATION|Exercise Examination archive with solutions (ALEEE)|1 Videos

  • SUPERPOSITION OF WAVES

    CHHAYA PUBLICATION|Exercise Entrance Corner (Integer answer type)|3 Videos

  • STATICS

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|3 Videos

  • THERMOMETRY

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIEVE - WBJEE|1 Videos

Similar Questions

Explore conceptually related problems

Equations of two light waves are y_(1) = 4 sin omega t and y_(2) = 3 sin (omega t + (pi)/(2)) . What is the amplitude of the resultant wave as they superpose on each other?

When the waves y_(1) = A sin omega t and y_(2) = A cos omega t are superposed, then resultant amplitude will be

If two waves represented by y_(1) =4 sin omega t and y_(2)=3"sin" (omega+(pi)/(3)) interfere at a point the amplitude of the resulting wave will about

Two waves represented as y_(1) = A_(1) sin omega t and y_(2) = A_(2) cos omega t superpose at a point in space. Find out the amplitude of the resultant wave at that point .

When two progressive waves y_(1) = 4 sin (2 x - 6t) and y_(2) = 3 sin (2x - 6t -(pi)/(2)) are superposed, the amplitude of the resultant wave is

The equations of two sound waves are given below : y_(1) = 0 . 2 "sin" (2pi)/(3) (330 t - x) m and y_(2) = 0 .02 "sin" (2pi)/(3 + k) (330 t - x) m If 10 beats are produced per sound due to superposition of these two waves, determine the value of k .

The equations of two sound waves are given gelow : y_(1) = 10 "sin" (pi)/(150) (33000 t - x ) cm and y_(2) = 10 "sin" (pi)/(165) (33000 t - x) cm How many beats will be produced per second due to superposition of these two waves ?

The equations for displacement of two light waves forming interference pattern are y_(1) = 4 sin omega t and y_(2) =3 sin (omega t+ (pi)/(2)) . Determine the amplitude of the resultant wave.

Two travelling waves superpose to form a stationary wave whose equation is y (x,t) = 5 sin (0. 1 pi x) cos 50 pi t where x, y are in cm and t is in x . Find the equations of the two superposing travelling waves .