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2 years ago

How does the doppler effect change wave speed?

1 answer:

4 0

It doesn't change wave speed at all.

It only changes the wavelength perceived by the observer
if he is moving toward or away from the source.

Notice that if one observer is moving toward a source, and another one
is moving away from the same source, then one observer would see an
abnormally long wavelength, and the other one would see an abnormally
short wavelength, both from the same source.

The Doppler Effect ... like beauty ... is in the eye of the observer.

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A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$