The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Where,

L = Vibrating length string

T = Tension in the string

$\mu =$ Linear mass density

At the same time we have the expression for the number of beats described as

$n = |f_1-f_2|$

Where

$f_1$ = First frequency

$f_2$ = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

$f \propto \sqrt{T}$

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

$n = f_2-f_1$

$f_2 = n+f_1$

Replacing $3/s$ for n and 202Hz for $f_1,$

$f_2 = 3/s + 202Hz$

$f_2 = 3/s(\frac{1Hz}{1/s})+202Hz$

$f_2 = 206Hz$

The frequency of the tightened is 205Hz

7 0

3 years ago

If a diffraction grating has 3700 lines per cm, what is the spacing d between lines

Sonja [21]

So first we find the gap between the slits by the formula d=1/N

N is number of lines per metre so 3700 line/cm = 370000 lines/m 
So d=2.7*10^-6 

Now we use the formula dsin(angle)=n(wavelength) 

d is the same 
n is the order of the diffraction pattern 

so wavelenth=dsin(angle)/n 
=[(2.7*10^-6)*sin30]/3 
=4.5*10^-7 m

7 0

2 years ago

In gym class you run 22 m horizontally, then climb a rope vertically for 6.2 m. What is the direction angle of your total displa

miskamm [114]

Im 99% sure
tan^-1 (6.2/22)= 15.7º
with 2 significant figures 16º

7 0

3 years ago

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