A tube closed at one end has a vibrating diaphragm at the other end , which may be assumed to be a displacement node . It is found that when the frequency of the diaphragm is `2000 Hz` , a stationary wave pattern is set up in which the distance between adjacent nodes is `8 cm`. When the frequency is gradually reduced , the stationary wave pattern reappears at a frequency of `1600 Hz`. Calculate
i. the speed of sound in air ,
ii. the distance between adjacent nodes at a frequency of `1600 Hz`,
iii. the distance between the diaphragm and the closed end ,
iv. the next lower frequencies at which stationary wave patterns will be obtained.

Explanation

To solve the given problem step by step, we will address each part of the question systematically.

### Given Data:
- Frequency at first observation, \( f_1 = 2000 \, \text{Hz} \)
- Distance between adjacent nodes at \( f_1 \), \( d_1 = 8 \, \text{cm} = 0.08 \, \text{m} \)
- Frequency at second observation, \( f_2 = 1600 \, \text{Hz} \)

### Part (i): Calculate the speed of sound in air
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