Physics, 07.03.2020 05:19 walkereddie580

A guitar string of length L = 0.94 m is oriented along the x-direction and under a tension of T = 92 N. The string is made of steel which has a density of rho = 7800 kg / m3. The radius of the string is r = 9.3 x 10-4 m. A transverse wave of amplitude A = 0.0020 m is formed on the string.
a. Calculate the mass per unit length u of the guitar string in kg/m.
b. Calculate the velocity (in m/s) of a traveling transverse wave on the guitar string.
c. Assume a form y) = A sin(a) for the transverse displacement of the string. Enter an expression for a of a transverse wave on a string traveling along the positive x-direction in terms of its wavenumber k, the position x, its angular frequency w, and the time t.
d. Assume a form y2 = A sin(a) for the transverse displacement of the string. Write an expression for a of a transverse wave on a string traveling along the negative x-direction in terms of its wavenumber k, the position x, its angular frequency w, and the time t.
e. Write an equation for a standing wave on the string y(x, t) created by y1(x, t) and y2(x, t) in terms of the amplitude of the original traveling waves A, its wavenumber k, the position x, its angular frequency w, and the time t. Use a trigonometric identity so that y(x, t) contains a sine term dependent only on k and x and a cosine term dependent only on w and t.

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Answer from: ihatemylifebutohwell

Part(a): The mass per unit length = $22.05 Kg m^{-1}$ .

Part(b): The velocity of the traveling transverse wave = $2.04 m s^{-1}$

Part(c): The expression for a transverse wave on the string travelling along positive x-direction is $y(x,t) = A \sin(kx - \omega t)$

Part(d): The expression for a transverse wave on the string travelling along negative x-direction is $y_{-}(x,t) = A \sin(kx - \omega t)$

Part(e): The equation of the standing wave is $y(x,t) = 2A \sin(kx) \cos(\omega t)$ .

Explanation:

Part(a):

The mass per unit length ( $\mu$ ) is given by,

$\mu = \pi~ r^{2} \rho = \pi \times 9 \times 10^{-4}~m~\times 7800~kg~m^{-3} = 22.05~Kg~m^{-1}$

Part(b):

According to the figure, the net force on the element ( $\Delta x$ ) of the string, is the sum of the tension in the string ( $F_{T}$ ) and the restoring force. The x-components of the force of tension will cancel each other, so the net force is equal to the sum of the y-components of the force. To obtain the y-components of the force, at $x = x_{1}$

$(\dfrac{\partial y}{\partial x})_{x = x_{1} }= \dfrac{- F_{V^{-}}}{F_{T}}$

and at $x = x_{2}$

$(\dfrac{\partial y}{\partial x})_{x = x_{2} }= \dfrac{F_{V^{+}}}{F_{T}}$

The net force ( $F_{net}$ ) on the string element is given by

$F_{net} = F_{V}^{+} + F_{V}^{-}\\~~~~~~~= F_{T}[ (\dfrac{\partial y}{\partial x})_{x = x_{2}} - (\dfrac{\partial y}{\partial x})_{x = x_{1}}]$

According to Newton's second law of motion, $F_{net} = \Delta x \times \mu$ , where $\mu$ is mass per unit length. So,

$&& \Delta x \times \mu \dfrac{\partial^{2}y}{\partial t^{2}} = [(\dfrac{\partial y}{\partial x})_{x = x_{2}} - (\dfrac{\partial y}{\partial x})_{x = x_{1}}] \times F_{T}\\&or,& \dfrac{\mu}{F_{T}} \dfrac{\partial^{2}y}{\partial t^{2}} = \dfrac{[(\dfrac{\partial y}{\partial x})_{x = x_{2}} - (\dfrac{\partial y}{\partial x})_{x = x_{1}}]}{\Delta x} = \dfrac{\partial^{2}y}{\partial x^{2}}$

Comparing the above equation with standard wave equation

$\dfrac{\partial^{2}y}{\partial x^{2}} = \dfrac{1}{v^{2}} \dfrac{\partial^{2}y}{\partial x^{2}}$

the the velocity () of the transverse wave is

$v = \sqrt{\dfrac{F_{T}}{\mu}} = \sqrt{\dfrac{92~N}{22.05~Kg~m_{-1}}} = 2.04~m~s^{-1}$

Part(c):

If the wave having wavelength ' $\lambda$ ' propagates along positive x-direction with the constant velocity '', then at any instant of time '' , then its angular displacement ' $\theta$ ' is related to the linear displacement 'x' is written as

$&& \dfrac{\theta}{x} = \dfrac{2 \pi}{\lambda}\\&or,& \theta = \dfrac{2 \pi}{\lambda} x$

Also the distance travelled by the wave at time 't' with constant velocity 'v' is .

So the wave equation can be written as

$y(x,t) = A \sin(\theta)\\~~~~~~~~~= A \sin(\dfrac{2\pi}{\lambda}(x - vt)) \\~~~~~~~~~= A\sin(\dfrac{2\pi}{\lambda}~x - \dfrac{2\pi}{\lambda} vt)\\~~~~~~~~~= A\sin(kx - \omega t)$

where '' is the wave number $k = \dfrac{2\pi}{\lambda}$ and $\omega = \dfrac{2\pi v}{\lambda} = \dfrac{2 \pi \dfrac{\lambda}{T}}{\lambda} = \dfrac{2 \pi}{T}$ is the natural angular frequency.

The wave equation propagating along positive x-direction is given by

. $y_{+} = A \sin(kx - \omega t)$

Part(d):

By the same argument given above The wave equation propagating along negative x-direction is given by

$y_{-} = A \sin(kx - \omega t)$

Part(e):

The wave equation propagating along positive x-direction is given by

$y_{1}(x,t) = A \sin(kx - \omega t)$

Similarly, the wave equation propagating along negative x-direction is given by

$y_{2} = A \sin(kx + \omega t)$

So the equation of the standing wave on the string created by $y_{1}(x,t)$ and $y_{2}(x,t)$ is the superposition of the waves and is given by

$y(x,t) = y_{1}(x,t) + y_{2}(x,t)\\~~~~~~~~~= A \sin(kx - \omega t) + A \sin(kx + \omega t)\\~~~~~~~~~= 2A \sin(kx) \cos(\omega t)$

A guitar string of length L = 0.94 m is oriented along the x-direction and under a tension of T = 92

Answer from: Quest

i dont know but

explanation:

whoever your teacher is, deserves a raise for using pokemon as an example

Answer from: Quest

answer: electrons

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