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.

Answers: 3

Show answers

Answers

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

The correct answer is c

Answer from: Quest

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

Other questions on the subject: Physics

Physics, 21.06.2019 22:00, natethawsm

Na food processing facility, a spherical container of inner radius r1 5 40 cm, outer radius r2 5 41 cm, and ther- mal conductivity k 5 1.5 w/m · k is used to store hot water and to keep it at 100°c at all times. to accomplish this, the outer surface of the container is wrapped with a 800-w electric strip heater and then insulated. the temperature of the inner surface of the container is observed to be nearly 120°c at all times. as- suming 10 percent of the heat generated in the heater is lost through the insulation, (a) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the container, (b) obtain a relation for the variation of temperature in the container material by solving the differential equation, and (c) evaluate the outer surface tempera- ture of the container. also determine how much water at 100°c this tank can supply steadily if the cold water enters at 20°c.

Answers: 2

continue

Physics, 22.06.2019 03:00, thebigman7993

The respiratory system removes oxygen and water from the body. select the best answer from the choices provided t f

Answers: 2

continue

Physics, 22.06.2019 05:30, trevorhenyan51

Suppose you have three polarizing filters, with the second at an angle of 42∘ to the first and the third at an angle of 90∘ to the first. by what perfect will the original intensity of unpolarized light be reduced to after passing through all three filters?

Answers: 2

continue

Physics, 22.06.2019 15:00, koranbutterton

Astudent throws a water balloon with speed v0 from a height h = 1.76 m at an angle θ = 21° above the horizontal toward a target on the ground. the target is located a horizontal distance d = 9.5 m from the student’s feet. assume that the balloon moves without air resistance. use a cartesian coordinate system with the origin at the balloon's initial position. (a) what is the position vector, rtarge t, that originates from the balloon's original position and terminates at the target? put this in terms of h and d, and represent it as a vector using i and j. (b) in terms of the variables in the problem, determine the time, t, after the launch it takes the balloon to reach the target. your answer should not include h. (c) create an expression for the balloon's vertical position as a function of time, y(t), in terms of t, vo, g, and θ. (d) determine the magnitude of the balloon's initial velocity, v0, in meters per second, by eliminating t from the previous two expressions.

Answers: 3

continue

You know the right answer?

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