Question

An organ pipe has one end closed and at the other end is a vibrating diaphragm, which is a displacement antinode.
When the frequency of diaphragm is 2,000Hz, a stationary wave pattern is set up in the tube. The distance between
adjacent nodes is 8.0 cm. As the frequency is slowly reduced, the stationary wave pattern disappears, but another
stationary wave pattern reappears at frequency 1,200Hz calculate the length of the tube.


Options:
a) 20cm
b)40cm
c)30cm
d)60cm

Answer 1

Explanation:

Given An organ pipe has one end closed and at the other end is a vibrating diaphragm, which is a displacement antinode. When the frequency of diaphragm is 2,000 Hz, a stationary wave pattern is set up in the tube. The distance between  adjacent nodes is 8.0 cm. As the frequency is slowly reduced, the stationary wave pattern disappears, but another  stationary wave pattern reappears at frequency 1,200 Hz calculate the length of the tube.  

• We need to find the length of the tube.
• So frequency is 2000 hz
• Distance between the nodes is 8 cm
• So in case of standing wave,
•            d = wavelength / 2
•           d = λ / 2  
•             = 0.08  
•         Now L = λ / 2
• To get a node it should be an integral multiple
•       So L = λ / 2 n where n belongs to N
•               = (0.08) n---------1
• And in another case we can see that
•             = (0.10) n’----------2
• Dividing 2 by 1 we get
•               n / n’ = 0.10 / 0.08
•                         = 5/4
• If n = 5 and n’ = 4, then
•           L = 0.08 x 5  
•             = 0.4
•            = 40 cm

Reference link will be

https://brainly.in/question/5217342