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General
Easy
Question
A source is moving across a circle given by the equation 
in with constant speed 
clockwise sense A detector is stationary at the point ( 2R, 0 ) w.r.t. the centre of the circle. The frequency emitted by the source is 
The coordinates of the source when the detector detects mimimum frequency is
- (R, 0)
- (0,-R)
The correct answer is: (0,-R)
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Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Two pulses are emitted by sources at S and | S . What is the time lag by which observer observe them?
Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Two pulses are emitted by sources at S and | S . What is the time lag by which observer observe them?
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Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Find the time 
Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Find the time
Physics-General
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Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Find the time 
Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Find the time
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A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
The maximum intensity produced at D is given by
A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
The maximum intensity produced at D is given by
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A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
If a minima is formed at the detector then, the magnitude of wavelength of the wave produced is given by
A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
If a minima is formed at the detector then, the magnitude of wavelength of the wave produced is given by
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A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
If a maxima is formed at a detector then, the magnitude of wavelength of the wave produced is given by
A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
If a maxima is formed at a detector then, the magnitude of wavelength of the wave produced is given by
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Superposition of waves results in maximum and minimum of intensities such as in case of standing waves. This phenomenon is called as interference. Another type of superposition result in interference in time which is called as beats. In this case waves are analyzed at a fixed point as a function of time. If the two waves are of nearby same frequency are superimposed, at a particular point, intensity of combined waves gives a periodic peak and fall. This phenomenon is beats. If w1 and w2 are the frequencies of two waves then by superimposed y = y1 + y2 , we get at x = 0, 
Thus amplitude frequency is small and fluctuates slowly. A beat i.e., a maximum of intensity occurs, also intensity depends on square of amplitude. The beat frequency is given by 
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Physics-General
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The figure shows the location of a source and detector at time t = 0. the source and detector are moving with velocities 
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Physics-General
