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3 years ago

12

What affect, if any, does increasing the speed of the plunger have on the wavelength of the waves

1 answer:

mojhsa [17]3 years ago

6 0

As the speed of the plunger increases, the wavelength of the waves decreases. The greater the frequency, the smaller the wavelength. The smaller the frequency, the greater the wavelength. When we increase the speed of the plunger, the frequency of the waves also increases, and just like with the size of the ball, it’s the speed of the plunger and the frequency of the waves are directly related.

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A car (mass = 1200 kg) is traveling at 31.1 m/s when it collides head-on with a sport utility vehicle (mass = 2830 kg) traveling

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4 years ago

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Two identical traveling waves, moving in the same direction, are out of phase by π/5.0 rad. What is the amplitude of the resulta

andreev551 [17]

Answer:

Therefore the amplitude of the resultant wave is $=0.95 y_m$

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The equation of wave:

y=A sin (kx-ωt)

For wave 1:

y₁=A sin (kx-ωt) = $y_{m}$ sin (kx-ωt)

For wave 2:

y₂=A sin (kx-ωt+Φ) = $y_{m}$ sin (kx-ωt+Φ)

Where A= amplitude= $y_m$

The angular frequency $\omega=\frac{2\pi}{T}$

$k=\frac{2\pi}{\lambda}$ , $\lambda$ = wave length.

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T= Time period

$\phi$ = phase difference = $\frac{\pi}{5}$

The resultant wave will be

y = y₁ + y₂

= $y_m$ sin (kx-ωt) + $y_m$ sin (kx-ωt+Φ)

$=y_m$ {sin (kx-ωt) + sin (kx-ωt+Φ)}

$=y_m\ sin(\frac{kx-\omega t +\phi + kx-\omega t }2)\ cos(\frac{kx-\omega t +\phi -kx+\omega t}2)$

$=y_m\ sin({kx-\omega t +\frac\phi 2)\ cos(\frac{\phi }2)$

$=y_m\ cos(\frac{\phi }2) sin({kx-\omega t +\frac\phi 2)$

Therefore the amplitude of the resultant wave is

$=y_m\ cos(\frac{\phi }2)$

$=y_m\ cos(\frac{\pi }{10})$

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6 0

4 years ago