Question

The number of vibrations n n per second of a nylon guitar string varies directly with the square root of the tension T T and inversely with the length L L of the string. If the tension is 256 256 kilograms when the number of vibrations per second is 15 15 and the length is 0.6 0.6 meters, find the tension when the length is 0.3 0.3 meters and the number of vibrations is 12 12 .

Answer 1

Answer: $T=40.96\ KgF$

Step-by-step explanation:

We know that:

$n$: The number of vibrations per second of the nylon guitar string.

$T$: Tension.

$L$: The length of the string.

Since  $n$  varies directly with the square root of the  $T$ and inversely with  $L$, the equation has the following form:

$n=k*\frac{\sqrt{T} }{L}$

Where "k" is the constant of variation.

Knowing that when $n=15$ and $L=0.6$, $T=256$, we can find the value of "k":

$n=k*\frac{\sqrt{T} }{L}\\\\L*n=k\sqrt{T}\\\\\frac{L*n}{\sqrt{T}}=k\\\\k=\frac{(0.6)(15)}{\sqrt{256}}\\\\k=0.5625$

Finally, in order to find the tension when the length is 0.3 meters and the number of vibrations is 12, you need to substitute these values and the value of "k" into $n=k*\frac{\sqrt{T} }{L}$ and solve for $T$:

$12=(0.5625)\frac{\sqrt{T} }{0.3}\\\\\frac{12(0.3)}{(0.5625)}=\sqrt{T}\\\\(6.4)^2=T\\\\T=40.96\ KgF$