the speed of light in glass is group of ___answer choices: same as in a vacuum. smaller than that in air. unchanged from that in air. larger than that in air.

A cylindrical component known as a piston oscillates in the cylinder head (cylindrical chamber) of a locomotive's engine. 0.76 meters is the stroke, which is twice the amplitude. Its greatest speed is 7.2m/s.

We use v[tex]_{m}[/tex] =ωx[tex]_{m}[/tex]=2πfx[tex]_{m}[/tex], where the frequency is  180/(60s)=3.0Hz  and the amplitude is half the stroke, or x[tex]_{m}[/tex]=0.38m. Thus,

v[tex]_{m}[/tex]=2π(3.0Hz)(0.38m)=7.2m/s.

A piston is a disc or short cylinder that fits snugly inside an engine cylinder and imparts motion to a pump or a liquid or gas by moving up and down against it. Pneumatic and hydraulic cylinders, reciprocating engines, reciprocating pumps, gas compressors, and other related devices all have pistons. It is the moving part that has a cylinder around it and piston rings that seal the gas tightness.

To transfer force from the expanding gas in the cylinder to the crankshaft via a piston rod and/or connecting rod is the function of a piston rod and/or connecting rod in an engine.

The processes of intake, compression, combustion, and exhaust are all four-stroke operations in automobile engines.

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The force applied on the body that is on the inclined plane is given as, F=mgsinθ, F=2×9.8×sin30  =9.8N and f=μmgcosθ f=0.7×2×9.8cos30 =11.88N.

Any increase in the moving force will result in slippage; limiting friction only exists when the moving force and the force opposing motion are equal.

The limiting frictional force is independent of the area of contact and proportional to the normal reaction between the contacting surfaces.

The amount of friction that can be produced when two static surfaces come into touch. A motion will start as soon as the force applied to the two surfaces surpasses the limiting friction. The normal response force and the coefficient of limiting friction are combined to get the limiting friction for two dry surfaces.

Therefore, The force applied on the body that is on the inclined plane is given as, F=mgsinθ, F=2×9.8×sin30  =9.8N and f=μmgcosθ f=0.7×2×9.8cos30 =11.88N.

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In terms of work produced, the gravitational force acting on block 2 is equal to that acting on block 1.

The quantity of labor is equal to zero when an object moves a fixed distance. The movement of the item as well as the applied load define the quantity of work. It rises in tandem with displacement and vice versa.

Blocks 2 and 1 are the same and are frictionlessly raised to the identical height, h = Z.

As a result, both blocks will have the same mass and gravitational pull.

m1 = m2

FG = m1g = m2g

Gravitational acceleration (g) is indicated here.

the labor put in while block 2 is subject to the gravitational pull.

W2 = m2gh

When gravity pulls on block 1, what labor is done on block 1?

W1 = m1gh

W1 = m1gZ

W1 = m2gZ , [m1=m2]

W1 = W2

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The question is -

Two identical blocks are pushed up frictionless inclines as shown. Consider the portion of the motion of block 1 during which it moves a distance of L 1 and the portion of the motion of block 2 during which it moves a distance of L 2. Each block is displaced the same vertical distance Z. The hand pushes parallel to the incline with the same magnitude of force F B H in both cases. Block 2 moves at a decreasing speed. Is the absolute value of the work done by the gravitational force on block 2 greater than, less than, or equal to the work done by the gravitational force on block 1? Explain.

This means that if object 1 hits object 2, object 2 will also hit object 1 with the same amount of force.

It is given that the string exerts a force on the arrow during the shooting of an arrow.

From Newton's third law, the arrow should also apply a force on the string. Thus, the reaction force is, arrow's push against the bowstring. Option (a) is correct.

The third law states that the two equal and opposite forces act on the two bodies in the system.

This means that if object 1 hits object 2, object 2 will also hit object 1 with the same amount of force.

The third law also satisfies the gravitational force between the two objects. The attractive force between the two objects is the same for both. Depending on their masses, their accelerations may vary, but the pull force between them remains the same.

Complete question:An archer shoots an arrow. Consider the action force to be the bowstring against the arrow. The reaction to this force is the:

A. arrow's push against the bowstring

B. weight of the arrow

C. air resistance against the bow

D. grip of the archers hand on the bow

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Answer:

Explanation:

The lower your center of gravity, the easier it is to keep your balance. If you're sitting on a chair, you can lean over more than if you're standing up. With your center of gravity low, you can lean further to one side or the other without creating enough turning force to tip you over.

The magnitude of the electrostatic force exerted on the proton by the electron is greater than that exerted on the electron by the proton.

What kind of force is electrostatic?

The Coulomb's inverse-square law, sometimes known as Coulomb's law, is an observational physical principle that measures the force exerted between two electrically charged particles that are stationary. Common names for the electric created between two charged objects at rest include electrostatic force and Coulomb force.

Between two charges that are separated by a distance, there is an electrostatic force. The size of each charges and the separation between them determine the strength of the electrostatic force. Two charges that are either positive or negative when placed together repel one another.

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The correct answers are: a) The rod is negatively charged and d) The rod is positively charged.

When the sphere is brought near the upper end of the rod and the rod rotates clockwise, this indicates that the sphere has a positive charge.

When the sphere is brought to point X near the lower end of the rod and the rod rotates counterclockwise, this also indicates that the sphere has a positive charge.

The other two options are not correct. The rod is not uncharged, because it has a charge when the sphere is brought near it. The rod also does not have no net charge until the sphere is brought near it, because it has a charge when the sphere is brought near it.

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The system's potential energy will be lost because it is no longer at rest, while its kinetic energy will increase.

Therefore,

Loss in potential energy = gain in (kinetic energy of the pulley + kinetic energy of the stone)                    ...........…..(1)

Let, the stone falls by a distance of h.

Given, the mass of the pulley disk mₚ = 2.50 kg

Its radius r = 20.0 cm = 0.2 m

Its moment of inertia I = (1/2)mₚr²

                                I = (1/2)(2.50 kg)(0.2 m)²

                                I = 0.05 kgm²

Given, the kinetic energy of the pulley disk = 4.50 J

If w the angular velocity of the pulley,

                               (1/2)Iw² = 4.50 J

                               (1/2)(0.05 kgm²)w²= 4.50 J

                               w² = 9/0.05 rad²/s²

                               w² =180 rad²/s²

                                v² = 180(0.2)^2 m²/s²

                               v² = 7.2 m²/s²

Given, the mass of the stone mₛ = 1.50 kg

Its kinetic energy = (1/2)(1.50 kg)(7.2 m²/s²) = 5.4 J

Potential energy loss of the stone = (1.50 kg)(9.80 m/s²)h

(a) Therefore, from equation (1), w get

      (1.50 kg)(9.80 m/s²)h = 4.5 J + 5.4 J

                                     h = (9.9 J)/(1.50 kg)(9.80 m/s²)

                                     h = 0.673 m

So, the stone must fall by a distance of  0.673 m.

(b) Total kinetic energy of the system = 9.9 J

The kinetic energy of the pulley = 4.50 J

The pulley’s fraction of kinetic energy = 4.50 J/9.9 J

                                                           = 0.455 = 45.5%

Therefore, the pulley has 45.5% of the total kinetic energy.

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The statement which is true regarding the motion of the satellite is D) The tangential velocity of the satellite does not remain constant.

Any object in orbit around the earth, the sun, or another large body is considered a satellite.

The satellite is able to maintain orbital motion even as gravity pulls it inward thanks to its tangential velocity. The necessary centripetal force is provided by gravitational attraction.

Newton was the first to say that a projectile launched with sufficient speed would actually orbit the earth.

The satellite moves faster when it is close to the earth and slows down while moving away to the top of the ellipse.

The question is incomplete. The complete question is ' A) The gravitational force exerted on the satellite by the moon will increase after time t0.

B) The satellite accelerates in a direction that is parallel to the direction of the tangential velocity.

C) The satellite will move toward the moon after time t0.

D) The tangential velocity of the satellite does not remain constant.'

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The couple moment M = 0.233 lb.ft

Given,

        The weight of disc A = 5 lb

        The radius of disc A = 0.5 ft

        The weight of disc B = 10 lb

        The radius of disc B = 0.75 ft

        Angular acceleration of disc A = 4 rad/s²

Let,

        The couple moment = M

Disc A:

         ∑[tex]M_{A} = I_{A} \alpha_{A}[/tex]

         M - F(0.5) = (1/2)(5/32.2)(0.5)²(4)

        2M - F = 0.078             ..........................(1)

Disc B:

          ∑[tex]M_{B} =I_{B} \alpha _{B}[/tex]

         F = (1/2)(10/32.2)(0.75)²[tex]\alpha _{B}[/tex]      .............(2)

         [tex]r_{A} \alpha _{A} =r_{B} \alpha _{B}[/tex]

        0.5(4) = (0.75)[tex]\alpha_{B}[/tex] .....................................(3)

By solving equations (1), (2), and (3) we get

         M = 0.233 lb.ft

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Conclusion/Discussion In conclusion, we tested the theory of Newton's Second Law of Motion using the Atwood Machine. In order to test the proportionalities the second law suggests, I was able to plot two graphs and successfully calculate the acceleration of each of my trials.

The second law of Newton will be thoroughly investigated in this experiment. According to Newton's second law, force is defined as mass times acceleration (F=mxa). Simply said, his law outlines the connection between an object's mass, acceleration, and required force to move it.

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The Mass of the Brown dwarf's star,M₂ = 4.132 * 10²⁷ kg

Let M₁ be the mass of the main-sequence star

M₂ be the Mass of the brown dwarf

v₁ be the speed of the main-sequence star

v₂ be the speed of the brown dwarf

The system's typical separation is 1 AU, where r₁ = r₂

The brown dwarf's orbital speed is 22 times that of the main-sequence star.

Therefore, v₂ = 22 v₁

As a main sequence star, the sun

The Mass of the main-sequence star is 2 * 10³⁰ kg

Centripetal force, F = Mv²/r

M₁v₁²/r₁² = M₂v₂²/r²

M₁v₁² = M₂v₂²

M₂/M₁ = v₁²/v₂²

M₂/M₁ = (1/22)²

M₂/M₁ = 0.002066

M₂ = 0.002066M₁

M₂ = 0.002066 * 2 * 10³⁰

M₂ = 4.132 * 10²⁷ kg

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In terms of the amount of light years energy received, star A should appear 100 times brighter than star B. This is due to the fact that received energy diminishes according to squared distance.

Contact binaries are binary star systems with members whose distance from one another is so great that their outer envelopes actually touch. This structure can actually be rather stable, with typical lifetimes measured in millions or even billions of years, which may surprise you.

When the Earth is at the other end of the star's orbit, six months later, astronomers take another measurement of the star's position. This results in a base line that is equal to the sum of the two angles used to measure the star and the double distance from Earth to the Sun's center (about 300,000,000 km). The distance to a nearby star can be calculated using these three variables in a very straightforward trigonometric manner.

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The position of the piston is 3.785 cm. The velocity of the piston is -2.585 cm/s. The acceleration of the piston is −15.14 cm/s The period of piston is 3.14 s and the amplitude of piston is 4 cm.

It is a kind of periodic motion bounded between two extreme points. For example, Oscillation of Simple Pendulum, Spring-Mass System. The object will keep on moving between two extreme points about a fixed point is called the mean position (or) equilibrium position along any path.

(a) The equation for the piston's position is given as

= x=(4.00 cm)cos(3t+6π )

At t=0,  

= x =(4.00 cm)cos(6π )

= x = 3.785 cm

(b) Differentiating the equation for position with respect to time gives us the piston's velocity:

= v = dt/dx = −(8.0 cm/s)sin(2t+ 6π )

At t=0,

= v = dt/dx = −(8.0 cm/s)sin(6π )

= v = -2.585 cm/s

(c) Differentiating again gives its acceleration:

= a = dv/ dt =−(16.0 cm/s 2 )cos(2t+ 6π )

At t=0,

= a = −(16.0 cm/s 2 )cos(6π )

= a = −15.14 cm/s

(d) The period of motion is = T

= T = ω/2π

= T = 2/2π

= T = 3.14 s

and,

The amplitude directly from the equation for x =

= A = 4 cm

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Answer:

Hope this helps.

Explanation:

(a) To calculate the astronaut's weight on Earth, you can use the formula for weight, which is W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2. So the astronaut's weight on Earth would be W = 90 kg * 9.8 m/s^2 = <<90*9.8=882>>882 N.

(b) The astronaut's mass when on the moon is the same as it is on Earth, which is 90 kg.

(c) To calculate the astronaut's weight on the moon, you can use the same formula as before, but with the acceleration due to gravity on the moon, which is 1.6 m/s^2. The astronaut's weight on the moon would be W = 90 kg * 1.6 m/s^2 = <<90*1.6=144>>144 N.

A specific sunspot can be followed as it moves over the Sun's surface as it revolves. This takes approximately from eastern to western limb.

While moving over the Earth-side of the sun, super sunspot AR2192 generated 10 major solar flares, six of which were X-class and four of which were above M5-class. These photos from NASA's SDO, taken between October 17 and October 29, 2014, show a sunspot that is the largest since November 1990 moving across the front of the sun. Try restarting your device if playback doesn't start right away. These photos from NASA's SDO, taken between October 17 and October 29, 2014, show a sunspot that is the largest since November 1990 moving across the front of the sun. Try restarting your device if playback doesn't start right away. Your watched videos might be added to.

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If the blade rotates with a constant angular velocity of 20 rad/s the work done on the blade by the motor in 1.0 min IS 960 J.

A torque is a force that causes rotation. Tells how effective force is at twisting or rotating an object.

Conditions for Equilibrium is that when FNet = 0 Translational EQ (Center of Mass) tNet = 0 Rotational EQ (True for any axis). Choose axis of rotation wisely to make problems easier, But as long as you’re consistent everything will be okay.

Gravitational Force Weight = mg, acts as force at center of mass, Torque about pivot due to gravity t = mgd, object not in static equilibrium.

As we know,

work = torque x angular displacement

θ = ωt

θ = 20 x 60

θ  = 1200rad

hence,

work = 0.80 x 1200

W = 960J

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The battery's emf is 8.87V and its internal resistance is 0.43. The internal resistance of the battery is 0.19, and its emf is 8.49 V.

1).

8.2=E−1.55r

10.40=E+3.52r

Solving these two equations we get

r=0.43Ω and E=8.87V

2).

8.2=E−1.54r

9.2=E+3.49r

Solving these two equations we get

r=0.19Ω and E=8.49V

The opposition a substance provides to the flow of electrical current is referred to as resistance. The capital letter R is used to symbolize it. The ohm, frequently written as a word and occasionally represented by the uppercase Greek letter omega, is the commonly used unit of resistance.

A component's resistance is one ohm when a one-volt potential difference (voltage) is present across it and an electric current of one ampere flows through it. (See Ohm's law for further information on how current, resistance, and voltage are related.)

Generally speaking, the current in a direct-current (DC) electrical circuit is inversely proportional to the resistance while the supplied voltage is held constant.

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4600 j work is done.

To find the amount of work done, the equation for work must be used. This equation is Work = Force x Distance. Force is equal to the mass of the box multiplied by the acceleration due to gravity, and the distance is 18 meters. The angle of 24 degrees above the horizontal must also be taken into account, therefore the cosine of 24 degrees must be used. The mass of the box is 64 kg and the acceleration due to gravity is 9.8 m/s2. After plugging in all the values, the work done is 4600 J.

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Angular speed when it is on the frictionless side of the track is 89.7 rad/s

Given that, mass, m=0.14 kg

Ball is released. from rest. at a height. of,  h= 0.83 m

Solid sphere of radius, R = 3.8 cm

                                       =0.038 m  

From the conservation of  energy

                          ΔK  = ΔU  

1/2 mv2 +   1/2 Iω2   = mgh

Here,  I=2/5MR2  ,  v=Rω

1/2 mv2 + 1/2 (2/5mR2)(v/R)2  =mgh

                          1/2 [v2 +2/5 v2] =gh

                           7/10 v2 = gh

                            0.7v2 = gh

                                  v =√ [gh / (0.7) ]

                                     =√ [(9.8 m/s2)(0.83 m) / (0.7) ]  

                                     = 3.408 m/s

Consequently, angular speed on the frictionless side of the rail,

                                  ω = v/R

                                     = (3.408 m/s)/(0.038 m)

                                     = 89.7 rad/s

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The points of the design altitude for the rocket, when the stagnation pressure is of 400 psi, using given plots will be :-

The Exit velocity and Pressure in the case of a rocket motor are 1290m/s and 8.1 atm respectively. The C* and thrust propellant calculations are based on certain formulas.

Therefore, these are the points of the design altitude for the rocket, when the stagnation pressure is of 400 psi, using given plots.

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[NOTE: THIS IS AN INCOMPLETE QUESTION. THE COMPLETE QUESTION IS: Consider a rocket motor equipped with a nozzle in which the stagnation conditions are 3000 K and 60 atm. The specific heat ratio is 1.25 and the molecular weight of the propellant is 18. The nozzle throat diameter is 5 cm and the nozzle exit diameter is 30 cm. a. Calculate the exit velocity b. Calculate the exit pressure c. Calculate c* for this propellant d. Calculate the mass flow rate for the rocket e. If the rocket is fired at sea level, calculate the thrust. f. If the rocket is fired at sea level, calculate the thrust coefficient g. If the rocket is fired at sea level, calculate the effective exhaust velocity h. If the rocket is fired at sea level, calculate the specific impulse. i. If the rocket is fired in space, calculate the thrust. j. If the rocket is fired in space, calculate the thrust coefficient k. If the rocket is fired in space, calculate the effective exhaust velocity I. If the rocket is fired in space, calculate the specific impulse]

The closed loop time constant , in seconds, is Vo(t) = (410/9 – 205/4 e-t + 205/36 e-9t) u(t).

What is steady state ?

The term "steady-state" refers to the condition in which a conductor can no longer absorb heat. During this stage, the temperature stabilizes. The heat, however, is constantly lost to the environment.

What is time constant ?

A first-order, linear time-invariant (LTI) system's reaction to a step input is characterized by a parameter known as the time constant, which is typically represented by the Greek letter tau. A first-order LTI system's primary defining attribute is its time constant.

Vin(s)     2/s+1     205/s+9     vo(s)

W(s) /Vin(s) = 2x205/(s+1) (s+9)

For step  response  

Vin(s) = 1/s

Vo (s) = 410/s (s) (s+9)

Using partial fraction method

Vo(s) = A/s + B/s+1 + C/s+9  - (1)

A= 410/(s+1) (s+9)|s=0  = 410/(0+1) (0+9)  = 410/9

A= 410/9

B= 410/s(s+9) |s=-1 = 410/(-1) (=1+9)  = -205/4

B = -205/4

C= 410/(s) (s+1)|s = -9  = 410/(-9) (-9+1) = 205/36

C=205/36

Substitute value of A,B and C in eq (1)

Vo(s) = 410/9s – 205/4(s+1) + 2052/36(s+9)

Apply inverse la place transform both side

L-1 (Vo(s) ) = L-1(410/9s) – L-1(205/4(s+1) ) + L-1 (205/36(s+9)

Vo(t) = 410/9 L-1 (1/s) -205/4 L-1 (1/s+1) + 205/36 L-1(1/(s+9))

Vo(t) =  410/9u(t) -205/4 e-t u(t) + 205/36 e-9t u (t)

Vo(t) = (410/9 – 205/4 e-t + 205/36 e-9t) u(t)

Therefore, the closed loop time constant , in seconds, is Vo(t) = (410/9 – 205/4 e-t + 205/36 e-9t) u(t).

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Since a metal ball is considered a solid sphere So the electric field inside the metal ball is given by E = (p/3∈₀)r

Where r is the distance from the center and rho is charge density. Hence at the center r = 0

Then electric field

E = (p/3∈₀) x 0 = 0 N/C

Hence electric field at the center of the metal ball is due to its own charge only being 0N/C.

An electrical property is associated with any point in space when some form of electric charge is present. The magnitude and direction of the electric field are represented by the value of E and are called field strength or electric field strength or simply electric field.

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The mirror should be held at a distance of 60cm

Magnification is defined as the ratio of the height of image to the height of object. It can also be defined as the ratio of distance of image from mirror or lens to the distance of object from lens.

Magnification is given as v/u . where v is the image distance and u is the object distance.

therefore 2 = v/u

v = 2u

from the mirror formula

1/f = 1/v+ 1/u . where F is the focal length

represent 2u for v

1/f = 1/2u + I/u

1/f = 3/2u

1/40 = 3/2u

2u = 120

divide both sides by 2

u = 120/2

u = 60cm

therefore the object distance is 60cm. This means that to have a double sized image , the distance from the mirror and his face must be 60cm.

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The tubes with lengths of 15, 45, and 75 cm are found to resonate with this wavelength. Additionally, resonance is seen for tubes with lengths of 25 and 75 cm and 31 and 93 cm.

For this reason, at the place where the tube is closed, we have a node, and at the open point, we have a belly; in this case, the fundamental wave is = 4L. To determine the length of the tube that has resonance, we must establish the natural frequencies of the tubes.

The following resonance, known as the first harmonic, is 3 = 4L / 3.

The next fifth harmonic resonance is 5 = 4L / 5, and we can see that its general form is n= 4L / n, where n = 1, 3, and 5.

Let's apply these terms to the issue at hand.

Start with the wavelength that is the shortest.

Lam equals 60 cm.

Let's determine the length of the tube this harmonica provides.

L = λ n / 4

n = 1 L = 60 will yield the shortest tube length. 1/4\s L = 15 cm

For n = 3 L = 60 3/4\s L = 45 cm

For n = 5 L = 60 5/4\s L = 75 cm

For n = 7 L = 60 7/4\s L = 105cm

We can see that the harmonics 1, 3 and 5 of the tubes with lengths 15, 45, and 75 resonate with this wavelength.

b) λ = 100 cm

For n = 1 L = 100 1/4\s L = 25 cm

For n = 3 L = 100 3/4\s L = 75 cm

For n = 5 L = 100 5/4\s L = 125 cm

There is resonance in the fundamental and third ammonium frequencies for lengths of 25 and 75 cm. L = 124 1/4 L = 31 cm

In the case of the second resonance, L = 124 3/4 L = 93 cm.

For tubes with lengths of 31 and 93 cm, resonance occurs in the fundamental and third harmonics.

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At point B, the gauge pressure is 22 kPa.

Calculation:

Given gauge pressure is 50 kPa.

Diameter is 3.0cm

speed is 2.0m/s

the gauge pressure is 22 kPa.

Pg denotes gauge pressure, which is related to absolute pressure as follows: Pa is the local atmospheric pressure, and pg is equal to p - pa. Example: 32.0 psi is the tire pressure as measured by an automobile tire gauge. 14.2 psi is the atmospheric pressure in the area.

In order to calculate gauge pressure, atmospheric (ambient) pressure is subtracted from absolute pressure. A scenario where the absolute pressure is higher than the atmospheric pressure is referred to as positive overpressure, and a situation where the absolute pressure is lower than the atmospheric pressure is referred to as negative overpressure.

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Incomplete question:

The 3.0-cm diameter water line shown at right splits into two 1.0-cm diameter pipes. All pipes are circular and at the same elevation. At point A, the water speed is 2.0 m/s and the gauge pressure is 50 kPa. What is the gauge pressure at point B?

A)12 kPa B)22 kPa C)29 kPa D)38 kPa E)42 kPa

The distance d between the doors, center to center is  11.3m

Location. of A = central. maximum, location. of B = first. minimum.

ΔY = [Ymin - Ymax] = λL/d (0 + 1/2) - 0 = 1/2 λL/d = 20m

distance d = λL/40m = 3 * 150/40 = 11.3m

The length of a particular path between two points, such as the distance walked through a maze, is what is referred to as an object's "distance traveled" [6]. Even a ball thrown straight up or the Earth after it completes an orbit can be considered as having a closed distance along a closed curve that begins and ends at the same location. As the curve's arc length, this is formalized mathematically.

A distance can also be signed, with a forward distance being positive and a backward distance being negative.

When constructing automobiles or mechanical gears, designers should take into account circular distance, which is the distance covered by a point on a wheel's circle (see also odometry). The wheel's circumference measures.

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Answer:

Solvent

Explanation:

In a solution of vinegar, water is the solvent. A solvent is a substance that dissolves another substance (in this case, acetic acid) to form a solution. The substance that is dissolved is called the solute. In the case of vinegar, the solute is acetic acid and the solvent is water.

Option C. None of the above choices are correct, A sound wave in air moves at a certain speed depending on its frequency. None of the wavelengths above it. all the aforementioned air temperatures. ambient temperature.

Under the same physical conditions, the speed of sound is constant for all frequencies in a particular medium. Since the distance covered by the sound wave in a given amount of time is what determines the speed of sound, it can be calculated using the following formula: where v is the velocity, is the sound wave's wavelength, and f is the frequency. Since the distance covered by the sound wave in a given amount of time determines the speed of sound, the following formula can be used to calculate it: Where v is the speed, is the sound wave's wavelength, and f is the frequency.

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The waves are called periodic wave if waveform repeats in time interval T.              A wave is said to be periodic if its duration of transmission is infinity. In a periodic wave wavelength and frequency is regulated. In a given time interval the wave completes one cycle.

For a periodic wave the wave function is y(x,t) = A sin2(x -vt) .The term 2 scales the wave to the natural period of the sin function. The term A gives the amplitude of the wave which is the maximum displacement of the wave.

If the velocity of the wave remains constant, then as the frequency of the wave is increased, the wavelength decreases  

The phase difference  between any two points on a wave is found as follows:-At P1 the phase is 1 = kx1- t1 +,  

At P2 the phase is 2 = kx2- t2 +.

So the phase difference  is  = 2-1 = k(x2 - x1)- (t2 - t1)                                                                   The wave speed is given by:-          v=k=2f2/=f    

Examples of periodic wave can be seen in:-

1.Electromagnetic wave in optical fibre

2.Sound from a guitar string

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