- #1
romanski007
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- Homework Statement
- The question was taken from the past exam paper.
Two rockets A and B are moving relative to each other. Rocket A emits a monochromatic signal of wavelength 600 nm every 1 hour ($$f_A$$ = 1 per hour), while rocket B receives them at a wavelength of 300 nm for the first part of the trip which took 6 hours from rocket A perspective.
The second part of the trip took 12 hours from rocket A perspective and rocket B observed a wavelength of 900 nm.
Determine the total number of signals received by rocket B for the entire duration of the trip.
Provided formula for the relative velocity of two rockets was $$v = c \frac{1-(\frac{\lambda_b} {\lambda_a})^2}{1+(\frac{\lambda_b} {\lambda_a})^2} $$
- Relevant Equations
- $$v = c \frac{1-(\frac{\lambda_b} {\lambda_a})^2}{1+(\frac{\lambda_b} {\lambda_a})^2} $$
I am assuming that B is a stationary observer here.
For the first part of the trip, using the formula, rocket A is approaching B at velocity $$v_A$$ =0.6 c.
The length that A travels is $$L_A = v_A t_1 $$ where $$t_1 = $$ 6 hrs. For the first part of the trip, B is receiving signals at a rate $$2 f_A$$ because the wavelength is halved and thus frequency is doubled.
The time taken for B to detect the turnaround of A is given by $$T_1 = \frac{L_1}{v_1} - \frac {L_1}{c} $$ where the first term is the length and the second term comes from the fact that A is approaching B.
Thus substituting, $$T_1 = 0.4t_1$$ and the total number of signals received $$N_1 = 2 f_A T_1 $$ = 4.8 signals.
For the second part, A is moving away with a velocity $$v_2 = \frac{5}{13} c $$ where the formula was used once again. The frequency is now given by $\frac{2}{3} f_A$ and the time of travel was $$ T_ 2 = \frac{L_2}{v_2} + \frac{L_2}{c}$$ where $$L_2 = v_2 t_2$$ and $$t_2$$ = 12 hours.
Furthermore second term emerges due to A moving away from B.
Thus total number of signals for second part is $$N_2 = \frac{2}{3} f_A T_2 = \frac{144}{13}$$ signals
Thus $$N = N_1 + N_2 = \frac{144}{13} + 4.8 = 16$$ signals
For the first part of the trip, using the formula, rocket A is approaching B at velocity $$v_A$$ =0.6 c.
The length that A travels is $$L_A = v_A t_1 $$ where $$t_1 = $$ 6 hrs. For the first part of the trip, B is receiving signals at a rate $$2 f_A$$ because the wavelength is halved and thus frequency is doubled.
The time taken for B to detect the turnaround of A is given by $$T_1 = \frac{L_1}{v_1} - \frac {L_1}{c} $$ where the first term is the length and the second term comes from the fact that A is approaching B.
Thus substituting, $$T_1 = 0.4t_1$$ and the total number of signals received $$N_1 = 2 f_A T_1 $$ = 4.8 signals.
For the second part, A is moving away with a velocity $$v_2 = \frac{5}{13} c $$ where the formula was used once again. The frequency is now given by $\frac{2}{3} f_A$ and the time of travel was $$ T_ 2 = \frac{L_2}{v_2} + \frac{L_2}{c}$$ where $$L_2 = v_2 t_2$$ and $$t_2$$ = 12 hours.
Furthermore second term emerges due to A moving away from B.
Thus total number of signals for second part is $$N_2 = \frac{2}{3} f_A T_2 = \frac{144}{13}$$ signals
Thus $$N = N_1 + N_2 = \frac{144}{13} + 4.8 = 16$$ signals