1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Oksi-84 [34.3K]
3 years ago
5

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 st

eel 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.

Physics
1 answer:
Ket [755]3 years ago
6 0

Answer:

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 (v) 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   'v', then at any instant of time 't' , 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 vt.

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 'k' 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)

You might be interested in
1. What are the 2 main categories of nonmetals?
Anettt [7]

the 2 main categories of nonmetals are REACTIVE NONMETALS an NOBLE GAS.

hope it helps

7 0
3 years ago
NewtonsXmeters / time is a unit of which of the following:
GuDViN [60]

Answer:

energy I think I'm not sure of the answer.

7 0
2 years ago
2. A Se lanza un electrón con rapidez inicial v0 = 1.60×106 m/s hacia el interior de un campo uniforme entre las placas paralela
Vanyuwa [196]

Answer:

A)     E = 145.6 N / C , B)  y= 2,8 10-7 m with a downward direction

C) he shape of the trajectory of the two particles is to simulate a parabola,

D)     F_{e} /F_{g} = 10³⁴

Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

           F = m a  

           - e E = m a

          a = - e E / m

with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

x axis (parallel to plates)

           x = v₀ t

           t = x / v₀

y axis (perpendicular to plates)

          y = y₀ + v_{oy} t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

          y = ½ a t²

we substitute

         y = ½ (e E /m)  (x / v₀)²

         y = ½ e x2 /m v₀²   E

we look for the electric field

        E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

         E = 2  9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

         E = 145.6 N / C

B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

          y = ½ e x² / m v₀² E

          y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)²   145.6

          y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

D) The force of gravity

           F_{g} = G m M / R²

electron

          Between the electron and the positive charges of the conducting plate

           F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²

           F_{g} = 4.1 10⁻⁵¹ N

           

electric force

           F_{e} = -e E

           F_{e} = - 1.6 10⁻¹⁹ 145.6

           F_{e} = 2.3 10⁻¹⁷ N

let's look for the reason between these two forces

         F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹

          F_{e} /F_{g} = 10³⁴

We see that the electric force is many orders of magnitude higher than the gravitational force.

5 0
3 years ago
The specific heat of fat is roughly 1,700 J/(kgoC) whereas that for water is around 4200 J/(kgoC). A 0.63 kg solution of lipids
BartSMP [9]

Answer:

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

Explanation:

Total heat content of the fat = heat content of water +heat content of the lipids

Let it be Q

the Q= (mcΔT)_lipids + (mcΔT)_water

total mass of fat  M= 0.63 Kg

Q= heat supplied = 100 W in 5 minutes

ΔT= 20°C

c_lipid= 1700J/(kgoC)

c_water= 4200J/(kgoC)

then,

100\times5\times60= m(1700)20+(0.63-m)(4200)20

solving the above equation we get

m= 0.46 kg

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

3 0
3 years ago
3. Why does any bonding occur (this includes ionic bonding and covalent bonding)?
Rzqust [24]

Answer:im not sure but hope this helps

Explanation:

Covalent bonds are formed because of sharing electrons whereas ionic bonds formation occurs because of transferring of electrons. Molecules are the particles in covalent bonds all through compound formation whereas in ionic bonds these are positively charged and negatively charged ions.

6 0
2 years ago
Other questions:
  • Certain mushrooms launch their spores by a catapult mechanism. As water condenses from the air onto a spore that is attached to
    15·1 answer
  • (20%) problem 1: suppose we want to calculate the moment of inertia of a 62.5 kg skater, relative to a vertical axis through the
    11·1 answer
  • A spinning rigid body has a rotational kinetic energy of 10 j and an angular speed of 10 s−1 . what is its rotational inertia?
    13·2 answers
  • r was thirsty and decided to mix up a pitcher of lemonade. She put lemon juice, water, and sugar into a pitcher and stirred it t
    15·2 answers
  • A lightning bolt occurs when billions of protons are transferred at the same time. ____________________
    13·1 answer
  • The blackbody curve of a star moving toward Earth would have its peak shifted (a) to lower intensity; (b) toward higher energies
    7·1 answer
  • If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?
    7·1 answer
  • If a boat and its riders have a mass of 1100 kg and the boat drifts in at 1.3 m/s how much work does Sam do to stop it
    8·1 answer
  • A marble, rolling with speed of 20m/sec rolls off the edge of the table that is 180m high (g=10m/sec2), find time taken to drop
    11·2 answers
  • for hundreds of years scientists deny the existence of rogue waves until the presence of when was finally caught on record. when
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!