Problem 05 e353 two identical audio speakers connected to...

Problem 05 (E35.3) Two identical audio speakers connected to the same amplifier produce in-phase sound waves with a single frequency that can be varied between 300 and 600 Hz. The speed of sound is 340 m s. You find that where you are standing, you hear minimum-intensity sound. (a) Explain why you hear minimum-intensity sound. (b) If one of the speakers is moved 39.8 cm toward you, the sound you hear has maximum intensity. What is the frequency of the sound? (c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?

Transcript

00:01 So here in this problem we are given that two identical speakers connected to the same amplifier and the frequency range of the amplifier is between 300 to 600 hertz and there are three parts in the equation in the a part we have to tell that you are standing somewhere near the speakers and you are able to hear the minimum intensity sound and we have to tell that why we are able to hear the minimum intensity sound so the reason behind this question is the concept behind the the concept behind the discussion is the destructive interference so destructive interference takes place when the waves came together in such a manner that they completely cancel out each other and for destructive interference for destructive interference to occur the path difference for the two sources must be a half integral number of wavelength that means r2 minus r1 this is the part difference it should be equals to m plus 1 divide by 2 times lambda so the point where you are standing should satisfy the condition for destructive interference to hear the minimum intensity so this was the solution for the part a now in the part b they are saying that one of the speaker is moved 39 .8 centimeters so let me write this any of the speaker is moved 309 .8 centimeters so let me write this any of the speaker is moved 30 and we are able to hear the maximum intensity sound and we will tell that what is the frequency for this sound so here we are able to hear the maximum sound intensity and it indicates that the speaker is moved at a distance equal to half wavelength from the previous position to satisfy the condition of constructive interference and the condition for constructive interference is r2 minus r1 is equals to m times lambda and lambda by 2 for the given condition should be equals to 39 .8 centimeters from here lambda will be equals to 79 .6 centimeter and if we get the answer in meters we will get 0 .796 meter now the equation for the frequency we know it is equal to frequency is equal to v over the lambda that is a wavelength this v is a speed of light and speed of light we know it is 340 meters per second and the value of lambda is 0 .796 meter so from here frequency will be equals to 427 hertz so therefore the frequency is equals to 427 hertz so this is the required answer for the b part now next i will be solving the c part.

03:21 Now in the c part we have to tell that how much closer to you from the position in the part b that we have calculated must the speaker be moved to the next position so that we can hear the maximum intensity sound...

Question

Please give Ace some feedback