Respuesta :
Answer:
beat frequency = 13.87 Hz
Explanation:
given data
lengths l = 2.00 m
linear mass density μ = 0.0065 kg/m
String A is under a tension T1 = 120.00 N
String B is under a tension T2 = 130.00 N
n = 10 mode
to find out
beat frequency
solution
we know here that length L is
L = n × [tex]\frac{ \lambda }{2}[/tex] ........1
so λ = [tex]\frac{2L}{10}[/tex]
and velocity is express as
V = [tex]\sqrt{\frac{T}{\mu } }[/tex] .................2
so
frequency for string A = f1 = [tex]\frac{V1}{\lambda}[/tex]
f1 = [tex]\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}[/tex]
f1 = [tex]\frac{10}{2L} \sqrt{\frac{T1}{\mu } }[/tex]
and
f2 = [tex]\frac{10}{2L} \sqrt{\frac{T2}{\mu } }[/tex]
so
beat frequency is = f2 - f1
put here value
beat frequency = [tex]\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}[/tex] - [tex]\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }[/tex]
beat frequency = 13.87 Hz
