Concepts Of Physics MCQ Edition [Volume 1]
- Exam
- Concepts Of Physics MCQ Edition [Volume 1]
- Subject
- Physics
- Chapter
- Sound Waves
Concepts Of Physics MCQ Edition [Volume 1] · Physics · Sound Waves
Two speakers S_1 and S_2 , powered by a common amplifier, are...
Two speakers and , powered by a common amplifier, are situated at and as shown in the figure. These speakers vibrate in phase at . A man is standing at a point on the X-axis located at a very large distance from the origin and begins to move parallel to the Y-axis. The speed of sound in air is given as . At what angle will the man experience a maximum of sound intensity for the first time?


Options
- A16^{}^{{}^{\circ}}
- B24^{}^{{}^{\circ}}
- C33^{}^{{}^{\circ}}
- D7.9^{}^{{}^{\circ}}
Correct answer
A. 16^{}^{{}^{\circ}}
Step-by-step solution
The distance between the two speakers and is .
The wavelength of the sound wave is given by:
For a point at a very large distance from the origin at an angle with the X-axis, the path difference between the sound waves reaching the man from and is:
The man starts from the X-axis (\theta = 0^{}^{{}^{\circ}}), where there is a central maximum. As he moves parallel to the Y-axis, he will experience the first maximum of sound intensity when the path difference equals one wavelength ().
Substituting the values, we get:
We know that \sin 15^{}^{{}^{\circ}} = \dfrac{\sqrt{3}-1}{2\sqrt{2}} \approx 0.259. Since is slightly greater than , the angle must be slightly greater than 15^{}^{{}^{\circ}}.
From the given options, the closest angle is 16^{}^{{}^{\circ}}.
The wavelength of the sound wave is given by:
For a point at a very large distance from the origin at an angle with the X-axis, the path difference between the sound waves reaching the man from and is:
The man starts from the X-axis (\theta = 0^{}^{{}^{\circ}}), where there is a central maximum. As he moves parallel to the Y-axis, he will experience the first maximum of sound intensity when the path difference equals one wavelength ().
Substituting the values, we get:
We know that \sin 15^{}^{{}^{\circ}} = \dfrac{\sqrt{3}-1}{2\sqrt{2}} \approx 0.259. Since is slightly greater than , the angle must be slightly greater than 15^{}^{{}^{\circ}}.
From the given options, the closest angle is 16^{}^{{}^{\circ}}.