Question

A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the fundamental frequency of the string is sounded with a 196.0-hz tuning fork, what beat frequency is heard?

Answer 1

First of all, we need to find the fundamental frequency of the string.

The fundamental frequency of a string is given by:
$f_1 = \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where L is the length of the string, T the tension and $\mu$ the linear density.
Using the information given in the exercise: L=0.660 m, T=56.7 N and $\mu =8.30 \cdot 10^{-4} kg/m$, we find
$f_1 = \frac{1}{2\cdot 0.660 m} \sqrt{ \frac{56.7 N}{8.30 \cdot 10^{-4}kg/m} }=198 Hz$

The beat frequency is given by the difference in frequency between the fundamental frequency of the string and the tuning fork (196 Hz), so it is:
$f_b = 198 Hz - 196 Hz = 2 Hz$

Answer 2

Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is 

$f = \sqrt{ \frac{T}{4mL} }$  -- (A)

here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg 


Plug in the values in Equation (A)

so $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ = 197.97Hz 

the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i