A shower stall has dimensions 86.0cm x 86.0 cm x 210 cm. If you were singing in

Question

A shower stall has dimensions 86.0cm x 86.0 cm x 210 cm. If you were singing in this shower, which frequencies would sound the richest (because of resonance)? Assume that the shower stall acts as a pipe closed at both emds, with nodes at opposite sides, and that the voices of various singers range from 130 Hz to 2000 Hz. Let the speed of sound in the hot air be 355 m/s.




Can someone clarify in step two where they're getting that the frequencies go from 1 to 9 and from 2 to 23? I have linked the solution. Or if you could possible provide a more coherent explanation as to how to go about solving this problem that would be appreciated as well.

Explanation / Answer

concept: when a pipe closed at both ends , we have a relation between the wave speed v frquency f , range of values n as given by   L = n v /(2f)
thus, frquency f = n v/ 2 L for , lenght , L = 86.cm = 0.86 m velocity of the sound ,v= 355 m/s The frequencies for different values of  n which fit the range of tones:
for , n=1, f = n v/ 2 L =206.39 Hz
for ,        n=2, f= 412.79 Hz
for ,        n=3, f= 619.18 Hz
for ,        n=4, f= 825.58 Hz
for ,       n=5, f= 1031.97 Hz
...
for ,       n=10, f=2063.95 Hz    ------(1) (Highest across the shorter distance)

for , lenght , L = 210 cm = 2.1 m The frequencies for  for different values of  n which fit the range of tones for , n=2, f = n v/ 2 L=169.04 Hz
for , n=3, f= 253.57 Hz
for , n=4, f= 338.09Hz
for , n=5, f= 398.80Hz
for , n=6, f=507.14 Hz
...
for , n=24, f =2028.57 Hz   ------(2) (Highest across the longer distance) thus, from equation (1) and (2)    height across the long distance nearlly cioncides thus, in equation (1) & (2)  voices of various singers range from 130 Hz to 2000 Hz is possibel only when 1 to   9     2 to 23 when a pipe closed at both ends , we have a relation between the wave speed v frquency f , range of values n as given by   L = n v /(2f)
thus, frquency f = n v/ 2 L for , lenght , L = 86.cm = 0.86 m velocity of the sound ,v= 355 m/s The frequencies for different values of  n which fit the range of tones:
for , n=1, f = n v/ 2 L =206.39 Hz
for ,        n=2, f= 412.79 Hz
for ,        n=3, f= 619.18 Hz
for ,        n=4, f= 825.58 Hz
for ,       n=5, f= 1031.97 Hz
...
for ,       n=10, f=2063.95 Hz    ------(1) (Highest across the shorter distance)

for , lenght , L = 210 cm = 2.1 m The frequencies for  for different values of  n which fit the range of tones for , n=2, f = n v/ 2 L=169.04 Hz
for , n=3, f= 253.57 Hz
for , n=4, f= 338.09Hz
for , n=5, f= 398.80Hz
for , n=6, f=507.14 Hz
...
for , n=24, f =2028.57 Hz   ------(2) (Highest across the longer distance) thus, from equation (1) and (2)    height across the long distance nearlly cioncides thus, in equation (1) & (2)  voices of various singers range from 130 Hz to 2000 Hz is possibel only when 1 to   9     2 to 23

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