Question

Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A is under a tension of 120.00 N. String B is under a tension of 130.00 N. They are each plucked and produce sound at the n=10 mode. What is the beat frequency?

Answer 1

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 × $\frac{ \lambda }{2}$      ........1

so  λ = $\frac{2L}{10}$  

and velocity is express as

V = $\sqrt{\frac{T}{\mu } }$    .................2

so

frequency for string A = f1 = $\frac{V1}{\lambda}$

f1 = $\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}$

f1 = $\frac{10}{2L} \sqrt{\frac{T1}{\mu } }$      

and

f2 = $\frac{10}{2L} \sqrt{\frac{T2}{\mu } }$

so

beat frequency is = f2 - f1

put here value

beat frequency = $\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}$  - $\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }$

beat frequency = 13.87 Hz