00:01
Hi, here this problem, it includes several small questions which are based upon the resonance of air column in a closed and organ pipe.
00:13
In the first question, using fundamental frequency of closed and organ pipe, which is given as n is equal to v, speed of sound here, divided by 4l, l is the length of that organ pipe.
00:29
Frequency this is given as 30 hertz here speed of sound this is 3993 meter per second divided by four times of the length of the tube length of the pipe which is missing so that length will be given by 393 divided by 4 into 30 so it comes out to be equal to 3 .275 meter or rounding it off up to 2 decimal places it becomes 3 .28 meter.
01:09
So this is the option e which is correct here.
01:17
Then in the second question, speed of sound is given as symbolically it is given as v.
01:31
So for the closed and organ pipe, frequency of the first harmonic means fundamental frequency will be given by n is equal to or let it be represented by f as it is frequency.
02:01
So this is f is equal to v by 4l, the fundamental frequency.
02:08
So here this is the option d which seems to be correct.
02:16
Then in the third question for resonance length of chamber in the closed and organ pipe, length of air chamber.
02:42
That should be equal to odd multiple of quarter of wavelength, one quarter of of wavelength because if we show the closed and organ pipe here this is one of the closed and organ pipe so when the air chamber will vibrate in its smallest frequency then there will be one -fourth of the wavelength being formed here half of the loop of a stationary wave so this half of the loop that will be equal to because a complete loop is equal to half of wavelength so half loop will be equal to lambda by 4 and another one if we vibrate the air column in the second possible frequency always there will be an antinode at the open end and a node at the closed end so second possibility will be this here this so here this is lambda by 4...