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Anna71 [15]
3 years ago
5

How can you double the frequency of a wave if you have control over both the wavelength and the wave velocity?

Physics
2 answers:
iren2701 [21]3 years ago
6 0

Answer:

Either double the speed or reduce the wavelength to half

Explanation:

We know that velocity of the wave is given by velocity=wavelength ×frequency

v=\lambda f

So f=\frac{velocity}{wavelength}=\frac{v}{\lambda }

As in the question it is given that we have control on both velocity and wavelength

So we have to double the the frequency then we can double the speed or reduce the wavelength to half

sammy [17]3 years ago
5 0

Answer:

Change the wavelength to half while keeping velocity constant or change the wave velocity to 2 times  while keeping wavelength constant.

Explanation:

The frequency of a wave can be defined as the rate of change of wave speed with respect to the wavelength of the wave.

Mathematically,

f=\frac{v}{\lambda}

Here, v is the velocity, lambda is the wavelength, and f is the frequency of the wave.

If the observer want to double the frequency of the wave, then he should increase the wave velocity two times and take wavelength constant for this case or either he should half the wavelength of the wave to take the velocity of the wave constant.

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A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

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Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

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In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic motion, with frequency 124 Hz
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Answer:

(a) A = 1 mm

(b) V_{max}=0.77872 m/s

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(c) The formula for the maximum acceleration is given by

a_{max}=\omega ^{2}A

a_{max}=778.72 ^{2}\times 0.001

[tex]a_{max}=606.4 m/s^{2}/tex]

8 0
3 years ago
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