Tag: superposition of waves-1: interference and beats

Two sound sources are moving in opposite direction with velocity $v _1$ and $v _2$ $(v _1>v _2)$. Both are moving away from a stationary observer.the frequency of both the source is $900\ Hz$. What is the value of $v _1 - v _2 $  so that the beat frequency observed will be $6\ Hz$  ?

Sound waves of equal amplitude wave frequency $(n-1),n _1(nH^1)$. They superinpose to give Beat. The no of Beats poduced per sec cuPll be:

A sources of sonic oscillations with frequency n= $1700$ Hz and a receiver are located on the same normal to a wall. Both the source and receiver are stationary, and the wall recedes from the source with velocity u= $6.0$ cm/s. Find the beat frequency registred by the receiver. The velocity of sound is equal to $v= 340$ m/s.

Two sound sources (of same frequency ) are placed at distance of 100 meter. An observer, when moving between both sources, hears 44 beats per second. The distance between sound source is now changed to 400 meter then the beats/second heard by observer will be  :

Two open organ pipes 80 and 81 cm long found to give 26 beats in 10 sec, when each is sounding its fundamental note. Then the velocity of sound in air is

Two monochromatic light waves of amplitudes $A$ and $2A$ interfering at a point, have a phase difference of ${60^0}.$ The intensity at that point will be  proportional to :

Two sound waves with wavelength $5$m and $5.5$ m respectively. each propoggate in a gas with velocity $300$ m/s. we expect the following number of beats per second

Two waves are approaching each other with a velocity of $16\, m/s$ and frequency $n$. the distance between two consecutive nodes is 

A body is walking away from a wall towards an observe at a speed of 1 m/s and blows a whistle whose frequency is 680 Hz. The number of beats heard by the observe per second is approximately.(velocity of sound in air = 340 m/s)

Two sound waves of equal intensity I produce beats . The maximum intensity of sound produced in beats will be