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2 years ago

A harmonic wave on a string is described by

Physics

1 answer:

xxMikexx [17]2 years ago

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$\huge{\bold{\orange{\underline{ Solution }}}}$

<h3>Given :-</h3>

A harmonic wave on a string is described by

$\sf{ Y( x, t) = 0.1 sin(300t + 0.01x + π/3)}$

x is in cm and t is in seconds

<h3>Answer 1 :-</h3>

Equation for travelling wave :-

$\sf{ Y( x, t) = Asin(ωt + kx + Φ)...eq(1)}$

Equation for stationary wave :-

$\sf{ Y( x, t) = Acos(ωt - kx )...eq(2)}$

Given equation for wave :-

$\sf{ Y( x, t) = 0.1 \:sin(300t + 0.01x + π/3)...eq(3)}$

On comparing eq(1) , (2) and (3)

We can conclude that, Given wave represent travelling wave.

<h3>Answer 2 :-</h3>

From solution 1 , We can say that,

$\sf{ Y( x, t) = 0.1 \: sin(300t + 0.01x + π/3).}$

It is travelling from right to left direction

Hence, The direction of its propagation is right to left that is towards +x direction.

<h3>Answer 3 :-</h3>

Here, We have to find the wave period

We know that,

Wave period = wavelength / velocity

Wave equation :-

$\sf{ Y( x, t) = 0.1 \:sin(300t + 0.01x + π/3).}$

ω = 300rad/s
k = 0.01

We know that, 

$\sf{v =}{\sf{\dfrac{ ω}{2π}}}{\sf{\: and\:}}{\sf{ λ =}}{\sf{\dfrac{ 2π}{k}}}$

Subsitute the required values,

$\sf{ wave\: period =}{\sf{\dfrac{ 2π/k}{ω/2π }}}$

$\sf{ wave \:period = }{\sf{\dfrac{k}{ω}}}$

$\sf{ wave\: period =}{\sf{\dfrac{ 0.01}{300}}}$

$\sf{ wave\: period = 0.000033\: s}$

<h3>Answer 4 :-</h3>

The wavelength of given wave

$\bold{ λ = }{\bold{\dfrac{2π}{k}}}$

Subsitute the required values, 

$\sf{ λ = }{\sf{\dfrac{2 × 3.14 }{0.01}}}$

$\sf{ λ = }{\sf{\dfrac{6.28}{0.01}}}$

$\sf{ λ = 628 \: cm }$

<h3>Answer 5 :-</h3>

We have wave equation

$\sf{ Y( x, t) = 0.1 sin(300t + 0.01x + π/3).}$

Travelling wave equation :-

$\sf{ Y( x, t) = A\:sin(ωt + kx + Φ)...eq(1)}$

Therefore, 

Amplitude of the wave particle

$\sf{ A = 0.1 \: cm}$

Hence, The amplitude of the particle is 0.1 cm

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Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

$K=\frac{1}{2}(0.5)(3)^2=2.25 J$

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

And about kinetic energy:

brainly.com/question/6536722

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