Most Important Selected Qs for JEE AdvancedPhysicsWaves and Sound
The vibrations of a string of length 600 cm fixed at both ends are represented by the equation y=4 ( x 15 ) (96 t) Where x and y are in cm and t in seconds
Options
- AThe maximum displacement of a point x =5 ~cm is 2 3 ~cm .
- BThe nodes located along the string are 15 n where integer n varies from 0 to 40 .
- CThe velocity of the particle at x =7.5 ~cm at t =0.25 sec is 0
- DThe equation of the component waves whose superposition gives the above wave are 2 2 ( x 30 +48 t ), 2 2 ( x 3
Correct answer
D. The equation of the component waves whose superposition gives the above wave are 2 2 ( x 30 +48 t ), 2 2 ( x 3
Step-by-step solution
(A) Displacement =4 Sin ( 5 15 )=4 Sin 5 =2 3 Cm (B) K x= 2 x= x 15 =30 Nodes wil form at 0, 2 , 2 2 , 3 2 , 4 2 , 5 2 .0,15,30,45 ie 15 is where n =0 to 40 (C) V= d y d t =-4 ( x 15 )(96 ) (96 t) at x=7.5 ~m ; t= 1 4 wv=-4 ( 75 15 )(96 ) (96 1 4 )=0 (D) 2 Sin ( x 15 +96 t ), 2 Sin ( x 15 -T 6 t )2 Sin 2 ( x 30 +48 t ), 2 Sin 2 ( x 30 -38 t )