InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Explain the formatin of stationary wave by analytical method. Show that nodes and antinodes are equally spaced in a stationary wave. The speed limit for a vehicle on road is 120 km/hr. A policeman detects a drop of 10 % in the pitch of horn of a car as it passes him. Is the policeman justified in punishing the car driver for crossing the speed limit ? ( Given : Velocity of sound = 340 m/s ) |
|
Answer» conditions for antinodes : when ` A = pm 2 a ` , the particles are these points vibrate with maximum amplitude which are called antinodes. We have ` A = 2a cos "" ( 2pi x ) / ( lamda) ` ` therefore pm 2 a = 2a cos "" ( 2 pi x ) /( lamda ) ` ` therefore cos "" ( 2pi x ) /( lamda) = pm 1 ` ` therefore ( 2pi x ) /( lamda ) = 0, pi, 2pi,...... ` ` therefore x = ( lamda p) /( 2 ) = p(( lamda ) /( 2 )) ` where p = 0, 1, 2 ............. For x = 0 , ` ( lamda) /( 2 ) , lamda , ( 3lamda) /( 2 ) `.... antinodes are prouduced. The distance between any two successive antinodes is ` ( lamda ) / ( 2 ) ` Condition for nodes : When A = 0 the particles at these points are permanently at rest which are called nodes. We have, ` A = 2a cos "" ( 2pi x ) / ( x ) ` ` 2a cos (( 2pi x ) /( lamda ) ) = 0 ` ` therefore cos (( 2pi x ) /( lamda ) ) = 0 ` `( 2 pi x ) /( lamda ) = (pi) / ( 2 ), ( 3pi ) /( 2 ), ( 5pi ) /(2) `.... ` x = ( 2P - 1 ) ( lamda)/( 4 ) ` Where P = 1, 2 , ............ For ` x = ( lamda ) /( 4 ) , ( 3lamda ) / ( 4 ), ( 5lamda ) / ( 4 ) .... ` nodes are produced The distance between any two successive nodes is ` ( lamda ) /( 2 ) ` The distance between any two adjacent antinode is ` ( lamda ) /( 4 ) `. Hence, in stationary waves, the nodes and antinodes are equispaced. Numerical : Given : ` " " n_ 2 = n _ 1 - 10 % n _ 1 , v= 340 m//s ` `" " n _ 2 = n _ 1 - ( 1)/( 10) n _ 1 = ( 9 ) /( 10 ) n _ 1 ` When the car approaches the policeman, apparent fequency is ` n _ 1 = ( v + v _ 0 ) /( v - v _s ) xx n " " `... (i) where ` v _ 0 = 0 ` and when the car passes the policeman apparent fequency is ` n _ 2 = ( v - v _ 0 )/( v + v _ s ) xx n " " ` ... (ii) where, ` v _ 0 = 0 ` and `v _ s ` is the speed of car. On dividing equation (i) by (ii) ` therefore " " ( n_ 1 ) /( n _ 2 ) = ( v + v _ s ) /( v - v _ s ) ` ` therefore ( n _ 1 )/((9 ) / ( 10 ) n _ 1 ) = ( 340 + v _ s ) /( 340 - v _ s ) ` ` therefore ( 10 )/( 9 ) = ( 340 + v _ s ) /( 340 - v _ s ) ` ` v _ s = 17.89 ` m/s ` = 64.42 km//hr ` ( speed of car ) The speed limit is 120 km/hr whereas the speed of car is 64.42 km/hr. Hence the policeman is not justified in his action. |
|