| Date | May 2011 | Marks available | 4 | Reference code | 11M.3.HL.TZ1.19 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Calculate and State | Question number | 19 | Adapted from | N/A |
Question
This question is about gravitational red-shift.
Two identical lasers are situated on the surface of the Earth. One is directed horizontally towards observer A, who measures the frequency to be 4.62×1014Hz. The other is directed vertically upwards towards observer B, who is at a height of 1.00×102m.
(i) State how the frequency as measured by observer B compares with the frequency as measured by observer A.
(ii) Calculate the difference in frequency between the laser light as measured by observers A and B.
(iii) State one assumption that you made in (a)(ii).
The lasers are now placed on a spaceship, which is accelerating upwards at a constant rate of 9.81ms−2, far away from any other masses as shown below. The distance of observer D from the laser is 1.00×102m. Observer C is at the bottom of the spaceship.
Explain, with reference to the equivalence principle, the frequencies measured by observers C and D, as compared to observers A and B.
Markscheme
(i) observer B measures a lower frequency;
(ii) \(\frac{{\Delta f}}{f} = \frac{{g\Delta h}}{{{c^2}}} \Rightarrow \Delta f = \frac{{4.62 \times {{10}^{14}} \times 9.81 \times 1.00 \times {{10}^2}}}{{{{\left[ {3.00 \times {{10}^8}} \right]}^2}}}\);
\({\Delta f}\)=5.04Hz;
Accept use of g = 10 ms−2 to get f = 5.13 Hz.
(iii) assume that g is constant over the height interval;
the equivalence principle states that it is impossible to distinguish between an accelerating reference frame and a gravitational field;
therefore the frequency measured by observer D will be lower than that measured by observer C (by 5.04Hz) / observer C measures the same value as A, observer D measures the same frequency as B / OWTTE;
