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natulia [17]
2 years ago
9

A plane wave has equation; y= 25sin(120 _4x).find the: (1)wave length (2)wave velocity (3)frequency and period of the wave

Physics
1 answer:
kupik [55]2 years ago
8 0

(1) The wavelength of the wave is 1.57 m.

(2) The velocity of the wave is 30 m/s .

(3) The frequency of the wave is 19.10 Hz and the time period of the    wave is 0.052 s.

Note: The given equation seems to be incomplete. Most probably the equation was y=25sin(120t-4x). And the standard system of units (i.e. kg, m, s) is used.

Wavelength: The distance between the consecutive crest or trough of a wave is called the wavelength of a wave.

The equation of a plane wave oscillating in the y direction and traveling in the x direction is given by the equation,

y=A sin(ωt-kx)

where y is the displacement along the y direction, A is the amplitude of the wave, ω is the angular frequency, t is the time, x is the displacement along the x direction, and k is the wave constant.

The given equation is,

y=25sin(120t-4x)

Comparing this equation with the above equation, following values are obtained.

k= 4 m^(-1)

ω =120 rad/s

A=25

The wavelength λ is given by the formula,

λ=2π/k

Here k = 4 m^(-1), so

λ=2π/4

λ=1.57 m.

Velocity: The velocity of a wave is the product of the frequency and the wavelength.

The velocity v is given by the formula,

v=ω*λ/2π

Here ω=120 rad/s and λ=1.57 m, so

v=120*1.57/2π

v=30 m/s

Frequency: The number of oscillations completed in one second is called the frequency of a wave.

The formula of the frequency is,

f=v/λ

Here v=30 m/s and λ=1.57 m, so

f=30/1.57

f=19.10 Hz

Time period: The time taken to complete one cycle of oscillation is called the time period of oscillation.

The formula to calculate the time period T is,

T=2π/ω

Here ω=120 rad/s, so

T=2π/120

T=0.052 s.

Learn more about velocity of wave.

brainly.com/question/25847786

#SPJ4

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4 0
3 years ago
A playground merry-go-round of radius R = 1.40 m has a moment of inertia I = 265 kg · m2 and is rotating at 11.0 rev/min about a
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Answer:

The value of new value of angular speed of merry go round.\omega_{2} = 0.96 \frac{rad}{sec}

Explanation:

Given data

r = 1.4 m

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N_{1} = 11 RPM

\omega_{1} = \frac{2 \pi N}{60}

\omega_{1} = \frac{2 \pi (11)}{60}

\omega_{1} = 1.15 \frac{rad}{sec}

From conservation of momentum principal

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265 × 1.15 = 317.92 × \omega_{2}

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Answer:

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The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

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For the y-direction gives

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Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

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Answer:

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