Answer:
Explanation:
From the given information:
The length of the shower stall L = 86.0 cm = 0.86 m
The width of the shower stall W = 210 cm = 2.1 m
Let assume that the stall acts as a closed pipe and the various singer voices range from 130 Hz to 2000 Hz.
Then; using wavelength equation;
[tex]\lambda = 2 L[/tex]
where;
[tex]permitted \ sounds \ L = ( \dfrac{n \lambda }{2})[/tex] &
[tex]v = f \lambda[/tex]
[tex]\lambda = \dfrac{v}{f}[/tex]
∴
[tex]L = \dfrac{nv}{2f} \ \ \ ... where, n = 1,2,3...[/tex]
By making the resonating frequency f the subject of the above formula, we have:
[tex]f_n = \dfrac{nv}{2L}[/tex]
where;
L = 0.860 m , then n = 1
Thus;
[tex]f_1 = \dfrac{355 \ m/s * 1}{2(0.860 \ m) }[/tex]
f = 206 Hz
For n = 9
[tex]f_9 = \dfrac{355 \ m/s * 9}{2(0.860 \ m) }[/tex]
[tex]f_9 = \dfrac{3195}{ 1.72}[/tex]
[tex]f_9 = 1857.56 \ Hz[/tex]
Thus, the resonant frequencies range for n = 1 to 9 with W = 2.10 m
However;
For n = 2
[tex]f_2 = \dfrac{355 \ m/s * 2}{2(2.10 \ m) }[/tex]
[tex]f_9 = 169 \ Hz[/tex]
For n = 23
[tex]f_{23} = \dfrac{355 \ m/s * 23 }{2(2.10 \ m) }[/tex]
[tex]f_{23} = 1944 \ Hz[/tex]
Thus, the resonant frequencies range from n = 2 to n= 23.
