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

beat frequency = 13.87 Hz

Step-by-step 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 ×
`( \lambda )/(2)` ........1

so λ =
`(2L)/(10)`

and velocity is express as

V =
`\sqrt{(T)/(\mu ) }` .................2

so

frequency for string A = f1 =
`(V1)/(\lambda)`

f1 =
`\frac{\sqrt{(T)/(\mu ) }}{(2L)/(10)}`

f1 =
`(10)/(2L) \sqrt{(T1)/(\mu ) }`

and

f2 =
`(10)/(2L) \sqrt{(T2)/(\mu ) }`

so

beat frequency is = f2 - f1

put here value

beat frequency =
`(10)/(2*2) \sqrt{(130)/(0.0065)}` -
`(10)/(2*2) \sqrt{(120)/(0.0065) }`

beat frequency = 13.87 Hz