Respuesta :
Answer:
[tex]\lambda=4L=1.33m[/tex]
v=343m/s
Explanation:
We have to take into account the expressions
[tex]f=\frac{2n+1}{4}\frac{v_s}{L}\\L=(2n+1)\frac{\lambda}{4}[/tex]
if we assume that 256Hz is the fundamental frequency we have
[tex]f=\frac{1}{4}\frac{v_s}{L}\\\\L=\frac{1}{4}\frac{v_s}{f}=\frac{1}{4}\frac{343\frac{m}{s}}{256Hz}=0.33m[/tex]
and for wavelength
[tex]\lambda=4L=1.33m[/tex]
hope this helps!!
Answer:
a) The wavelength, λ = 1.352 m
b) The speed of the sound in the column air is v = 346.112 m/s
Explanation:
a) Frequency, f = 256 Hz
The water level is lowered by 676 mm. i.e. x = 676 * 10⁻³ m
The length, x, at which the loudest sound will be heard is λ/2
i.e. x = λ/2
Therefore the wavelength of the 256 Hz sound waves in the column air is given by the relation
λ = 2x
λ = 2 * 676 * 10⁻³
λ = 1.352 m
b) The speed of the sound in the column air:
The speed, v, is given by the relation v = f λ
speed, v = 256 * 1.352
v = 346.112 m/s