Respuesta :
Answer:
a) [tex]\lambda=1\ m[/tex]
b) [tex]f=122.47\ Hz[/tex]
c) [tex]\lambda_s=2.8\ m[/tex]
Explanation:
Given:
distance between the fixed end of strings, [tex]l=1.5\ m[/tex]
mass of string, [tex]m=5\ g=0.005\ kg[/tex]
tension in the string, [tex]F_T=50\ N[/tex]
a)
Since the wave vibrating in the string is in third harmonic:
Therefore wavelength λ of the string:
[tex]l=1.5\lambda[/tex]
[tex]\lambda=\frac{1.5}{1.5}[/tex]
[tex]\lambda=1\ m[/tex]
b)
We know that the velocity of the wave in this case is given by:
[tex]v=\sqrt{\frac{F_T}{\mu} }[/tex]
where:
[tex]\mu=[/tex] linear mass density
[tex]v=\sqrt{\frac{50}{(\frac{m}{l}) } }[/tex]
[tex]v=\sqrt{\frac{50}{(\frac{0.005}{1.5}) } }[/tex]
[tex]v=122.47\ m.s^{-1}[/tex]
Now, frequency:
[tex]f=\frac{v}{\lambda}[/tex]
[tex]f=\frac{122.47}{1}[/tex]
[tex]f=122.47\ Hz[/tex]
c)
When the vibrations produce the sound of the same frequency:
[tex]f_s=122.47\ Hz[/tex]
Velocity of sound in air:
[tex]v_s=343\ m.s^{-1}[/tex]
Wavelength of the sound waves in air:
[tex]\lambda_s=\frac{v_s}{f_s}[/tex]
[tex]\lambda_s=2.8\ m[/tex]
