Physics, 30.03.2020 21:29 Maaarii

A guitar string is strummed near a tuning fork that has a frequency of 512 Hz. Initially, the guitar and tuning fork together create a sound wave with a beat frequency of 5 Hz. The tension in the guitar string is then increased, after which the guitar and tuning fork together create a sound wave with a beat frequency of 4 Hz. After the tension in the guitar string is increased, is the frequency of the guitar string:

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Answer from: YRNlightskin3360

The question is incomplete. However, I have made an attempt to guess what the question is asking for.

The original frequency of the guitar string is 507 Hz.

The frequency is increased after the tension is increased.

Its new frequency is 508 Hz.

Explanation:

The tuning fork has a frequency of 512 Hz.

With a beat frequency of 5 Hz, the guitar has a frequency of

512 - 5 = 507 Hz OR512 + 5 = 517 Hz

The relationship between frequency, f, and tension, T, is given by

$f\propto\sqrt{T}$ provided other factors are constant.

It follows that when tension is increased, frequency increases.

Therefore, when the tension in the string is increased, its frequency increases.

From the question, this increase in frequency caused a reduced beat frequency of 4 Hz.

This means the new frequency is

512 - 4 = 508 Hz OR512 + 4 = 516 Hz

In the first choice, there is an increase in the frequency from 507 Hz to 508 Hz.

The second choice indicates a decrease from 517 Hz to 516 Hz.

Since we have established there is an increase in frequency, the first choice is correct. The original frequency of the guitar string is 507 Hz.

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

Answer from: Quest

The purpose of a lab is to do your "expirements" the importance of the topic is so you can now what the story is about and the last one is because no one is me with my question ( what is 3 over 4 x - 9 = 27 maybe you can )

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A guitar string is strummed near a tuning fork that has a frequency of 512 Hz. Initially, the guitar...