For a certain organ pipe, three successive resonance frequencies are observed at 425, 595 and 765 Hz, respectively. The length of the pipe is (speed of sound in air = 340 m s - 1 )

Step-by-step solution

For closed organ pipe, resonant frequency is given by, ν n = 2 n + 1 V 4 l , where V & l represents speed of sound and length of organ respectively. For first case, 2 n + 1 V 4 l = 425     . . . 1 For second case, 2 n + 1 + 1 V 4 l = 595 ⇒ 2 n + 3 V 4 l = 595     . . . 2 On solving 1 & 2 , we have ⇒ 2 V 4 l = 170 ⇒ l = 2 × 340 4 × 170 = 1   m