missqtee78
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The displacement, d, in millimeters of a tuning fork as a function of time, t, in seconds can be modeled with the equation d = 0.6 sin (3520rt) . What is the frequency of the tuning fork?A,1/3520 Hz B. 1/1760 Hz C. 1760 Hz D. 3520 Hz

Respuesta :

Answer:

C is correct. 1760 Hz

Step-by-step explanation:

The displacement (d) in millimeters of a tuning fork as a function of time t

[tex]d=0.6\sin(3520\pi t)[/tex]

If we compare with wave equation

[tex]y=a\sin(\omega t)[/tex]

If we compare it with our equation we get

[tex]\omega=3520\pi[/tex]

As we know the formula of [tex]\omega=2\pi \nu[/tex]

[tex]\therefore 2\pi \nu=3520\pi[/tex]

[tex]\nu=1760\text{ Hz}[/tex]

Thus, The frequency of the tuning fork is 1760 Hz

ashishdwivedilVT

The frequency of the tuning fork using the wave equation is 1760Hz.

It is given that:

Displacement equation:

[tex]d = 0.6Sin 3520\pi t[/tex]..........Eq1

The general form of the wave equation is:

[tex]y=a Sin At[/tex].............Eq2

Where A is the angular frequency.

What is the formula of angular frequency in terms of natural frequency?

Angular frequency A = [tex]2\pi f[/tex]

Where f is natural frequency.

If we compare Eq1 and Eq2,

A = 3520π

[tex]2\pi f=3520\pi \\\\f =1760Hz[/tex]

Therefore, the frequency of the tuning fork using the wave equation is 1760Hz.

To get more about wave equation visit:

https://brainly.com/question/25699025

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