Question

13. 4/6 points | Previous Answers SerPSE10 17.6. P024 My Notes Ask Your A shower stall has dimensions 86.0 cm x 86.0 cm x 210 cm. Assume the stall acts as a pipe closed at both ends, with nodes at opposite sides. Assume singing voices range from 130 Hz to 2 000 Hz and let the speed of sound in the hot air be 355 m/s. For someone singing in this shower, which frequencies would sound the richest (because of resonance)? For L 0.860 m fn = n 1857.551× Hz for each n = 1 |v' to 9 For L' = 2.10 m to 23 for each n = fn' = n 1944-041x Hz Need Help? Read t Submit Answer Save Progress

Answer 1

Ans: For a open pipe,

a) Frequency of standing waves is given by

  

355n = 206 395n=nfl nu nL2x 0,860

  Here f₁ is fundamental frequency and f₁ = 206.395 Hz

The maximum frequency available is 2000 Hz. So 9f₁ = 1857.56 will be in the given voice range.

b) Frequency of standing waves is given by

  

nu 355n 2L2 × 2.10 84.524n = nfl n - -

  Here f₁ is fundamental frequency and f₁ = 84.524 Hz

The maximum frequency available is 2000 Hz. So 2f₁ = 169.048 and 23f₁ = 1944.048 will be in the given voice range.