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3 years ago

Why did the magnitude of the doppler shift increase as the velocity of the sound source increased?

Physics

2 answers:

geniusboy [140]3 years ago

7 0

Explanation:

Doppler effect: It is the phenomenon in which there is relative motion between the source and the observer or the source and the listener. There is an apparent change in the frequency.

When the observer and the source are moving towards each other then the frequency increases.

When the observer and the source are moving away from each other then the frequency decreases.

The expression of the speed of the wave in terms of the frequency is as follows;

$v=f\lambda$

Here, v is the speed of the wave, f is the frequency of the wave and $\lambda$ is the wavelength of the wave.

From the above relation, it can be concluded that speed of the sound is directly proportional to the frequency of the wave.

The magnitude of the Doppler shift increases when the velocity of the sound source is increased.

jolli1 [7]3 years ago

6 0

The Doppler effect happens as the result of a source of waves having relative motion to observers somewhere else consider that if something is producing a wave of wavelength W and it is moving relative to the waves then as the waves spread out it seems to chase one side and run away from the other side there for wavelengths produced on the scale it is chasing will be less than W on the side it is running away from the apparent wavelength will be greater than W so using this knowledge if the source of the Waves moves faster than it will chase one of the more closely and run away from the other more quickly there for making the short wavelength even shorter and the long wavelength even longer can turn you in and tell you have gone all over the speed of wave creating boom when dealing with sound this is what causes a sonic boom this lengthening of the long wavelengths and the shortening of the short wavelengths is referred to as being a greater Doppler shift

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Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

$F = G \frac{m_1m_2}{r^2}$

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

$F = m_2a$

Therefore, we can see that

$a(r) = G \frac{m_1}{r^2}$

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}$

a(R) is actually the constant acceleration at sea level

and

$a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}$

Therefore:

$a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})$

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}$

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3 years ago

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